MYSQL :: Outer Join Syntax Error

Left Outer Join Syntax
I am trying to join a table to three other tables

Table 1 has all records
Table 2-4 must be left outer joined to table 1, not in a chain

using "other" dbs ...

select t1.a
,t2.b
,t3.c
,t4.d

from t1
,t2
,t3
,t4

Where t1.id = t2.id (+) and t1.id = t3.id (+) and t1.id = t4.id (+)

I've tried using the left outer join syntax ...

select t1.a
,t2.b
,t3.c
,t4.d

From t1
LEFT OUTER JOIN t2 using (id)
LEFT OUTER JOIN t3 using (id)
LEFT OUTER JOIN t4 using (id)

but that, of course chains t1 to t2, t2 to t3, and t3 to t4.

How can I make this join work, without the use of subqueries?

Assitance Requested On Syntax With Joins (outer)
Select d.* from
( select agent_addr, max(date_time) from eventlog where priority = 1
group by agent_addr) d,
(select agent_addr, max(date_time) from eventlog where priority = 6
group by agent_addr) u
where d.agent_addr = u.agent_addr
and d.access_time >= u.access_time

The above statement is to get certain data from an event table. I need
all of the lines where the event is 1 and there are no newer event 1's
or event 6's

I need to know what devices are currently down (Priority 1) based on
this history table. where there is no newer event 1 entries for the
agent_addr, or there is no newer priority 6's for that agent_addr
(priority 6 means device is back up)

I'm stumped on the syntax to do joins, or to get this to work.

the table works like the following
device goes down entry is made as priority 1 eventid 1
device comes back up new endtry is made priority 6 eventid 2
device goes back down another entry is made as priority 1. eventid 3

I need to make sure i only am getting a list of the latest status of
each device in the aformentioned scenario it would be the one with
event ID 3

Outer Join With Where
I am willing to select all the distinct Products from the product table and in case the PC Name was the same as a @ passed PC name return ALSO the local path, Otherwise return null for the localpath field.... How I can do that?
here is my query:

SQL
SELECT Products.Name, Products.Code,
UserProduct.LocalPath, UserProduct.PC FROM
Products LEFT OUTER JOIN
UserProduct ON
UserProduct.ProdCode = Products.Code
this returns the same product name and product code from many users....also, I want to return distinct Product Names also.

Do I Need An Outer Join?
I have a table that stores traffic data keyed on a char (even though I call it a tstamp). The data is coalated into hourly chunks and is viewed through a web app in monthly, daily and hourly increments.

I've been charged with creating some KPI's (Key Performance Indicators) from this data and one of them is for a given month/day/year, generate an addition figure of how many times the average traffic per period (in a column called average) is less than or equal to 5.

mysql> desc coal_counts ....

Outer Join
hello, I am trying to run following query in mysql.

select a.name, a.record, b.data from A a, B b where a.name*=b.name and b.date='2006-07-28'

How can I do this in mysql? I looked up LEFT JOIN but still can't figure it out.

OUTER JOIN
I have 2 tables with the corresponding columns:

1) InventoryComputers-ComputerID, ComputerSN.
2) InventoryStations-Autonum, BranchID, StationID, ComputerID.

I want to display all rows of InventoryComputers for which the ComputerID does not have an entry in InventoryStations.
So basically, I want to display those computers that do not have a station assigned to them.

I tried this sql statement, but it also displays those rows that are present in InventoryStations table, which I don't want.

SELECT * FROM InventoryComputers
LEFT OUTER JOIN InventoryStations on InventoryStations.ComputerID=InventoryComputers.ComputerID
ORDER BY InventoryComputers.ComputerID

Any ideas what is wrong??

Outer Join
i have a main table called property, and a lookup table called ptype (property type). the propertry table has a "type" field, and a "subtype" field. both these fields look at the ptype table to get their titles.

so, i currently have the query below, but i also want to add to this the subtype. trouble is the field name is going to be the same for both the type and subtype values, so how do i get at them both?

I tried "AS subtype" after the LEFT OUTER JOIN but this didnt work.

SELECT * FROM property
LEFT OUTER JOIN ptype ON ptype.pty_id = property.pro_ptype
WHERE property.pro_id = 1

LEFT OUTER JOIN + WHERE ?
I have problem with constructing a SQL query. I have tables with data:

mysql> SELECT * FROM category;

+----+-----------+
| id | is_active |
+----+-----------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
+----+-----------+

4 rows in set (0.00 sec)

mysql> SELECT * FROM category_i18n;

+------------------+----+---------+
| name | id | culture |
+------------------+----+---------+
| Agroturystyka | 1 | pl_PL |
| Bed&breakfast | 2 | pl_PL |
| Hotel | 3 | pl_PL |
| Obiekt zabytkowy | 4 | pl_PL |
+------------------+----+---------+

4 rows in set (0.00 sec)
Category table includes category id and is_active status. Category_i18n has category names translations for several cultures and the id field is a foreign key.

Now I'd like to get result like this:
mysql> SELECT c.id, ci.name FROM category c LEFT OUTER JOIN category_i18n ci ON c.id=ci.id WHERE ci.culture='pl_PL' OR ci.culture IS NULL;

+----+------------------+
| id | name |
+----+------------------+
| 1 | Agroturystyka |
| 2 | Bed&breakfast |
| 3 | Hotel |
| 4 | Obiekt zabytkowy |
+----+------------------+

4 rows in set (0.00 sec)
And in this case (culture='pl_PL') it works fine. But when I change culture to 'en_US' I don't get any results. How to change my query to select all ids from category table and join name field from category_i18n table but with culture condition. If there is no translation there should be a NULL value.

Result I'd like to get for culture='en_US':

+----+------------------+
| id | name |
+----+------------------+
| 1 | NULL |
| 2 | NULL |
| 3 | NULL |
| 4 | NULL |
+----+------------------+

Left Outer Join
I have noticed when I do a Left Outer Join in short form that many rows
become missing as result of null values.

i.e Left Outer Join (t1,t2,t3) ON (....)

T3 being table with null values in

It seems to be all to hard when I log a bug.

Three Table Outer Join
I have three tables:

items (id, article, etc)
deliveries (id, shipment_date, etc)
movements (item_id, delivery_id, quantity, etc)

I want to get result as following:
article | sum(quantity)

but for all articles (outer join) with condition for deliveries table(shipment_date >= '2004-08-01' for example)

It works for two table join (items and movements) without deliveries:
select i.id, sum(m.quantity) from
items i left outer join movements m on m.item_id = i.id

but how I should join "deliveries" table to add condition for shipment_date ?

Left Outer Join
I have just converted an app from MS Access to MySql. I have using Hyperion Performance to run my queries. I have 4 tables and main table and three look up tables. when I do a left outer join to one table the query runs fine, however when I do the left outer join to more then one table it error out.

with this error message: You have an error in your sql syntax check the manual that corresponds to your Mysql server version for the right syntax to use near {oj left outer join rc_availability Al4 Al1.rc=Al4.rc} where al1.msa=al at line 1 Code:

Left Outer Join
This is what I have come up with:

SELECT CP1.PosID, CP1.Rank, CP1.BookID, CP2.BookID AS LastWeekBook, CP1.StationID, CP1.ChartTypeID, CP1.WeekID ,
CP2.Rank AS LastWeekRank, CP2.WeekID AS LastWeek
FROM Position AS CP1
LEFT OUTER JOIN Position AS CP2 ON CP1.BookID = CP2.BookID
WHERE CP1.WeekID = {d '2007-05-08'}
AND CP2.WeekID = DATE_SUB({d '2007-05-08'}, INTERVAL 7 DAY)
AND CP1.ChartTypeID = 101
AND CP2.ChartTypeID = 101
ORDER BY Rank;

(Note: joins to other tables containing author infor and book titles etc, have been omitted for simplicity.)

Instead of returning 50 rows (one for each matching row in CP1) it returns 36 rows (only those rows where the book appears in both CP1 and CP2).

I have tried alternative join syntax as follows and all return the same result:
LEFT JOIN
INNER JOIN (36 rows would be expected)
RIGHT JOIN
RIGHT OUTER JOIN

As far as I can tell the SQL as written should be giving me the desired result but it isn't and I can't see what I might be doing wrong.

Iso FULL OUTER JOIN
I need to do a full outer join, but have not seen this available in the fine manual. Is this type of join available?

The task is to list everyone that has departed the company in a certain date range. Sometimes a record is entered on table A, sometimes on table B, sometimes on both.

SELECT a.NAME, b.NAME FROM TABLEA A FULL OUTER JOIN TABLEB B ON A.SSN = B.SSN WHERE A.DEPART_DATE BETWEEN @BEGIN AND @END or
B.DEPART_DATE BETWEEN @BEGIN AND @END .

Outer Join And Distinct
I just started building a website and use MySQL as my backend database. I'm having a little trouble creating an SQL query that will get the correct information out.

Here's my problem:

I have 3 tables:
people: stores the name of people who can vote
votetype: stores the type of votes there are (In favor, Against, Withheld)
decisions: the dicisions that were voted on

Then I have another table that links them all together called votes. Votes has three columns: personID, typeID and decisionID. personID and decisionID are primary key. All are foreign key to obvious tables.

I want to retrieve for a given person the votes he made for all decisions. So: a list of all decisions with the vote he or she made and a NULL if there is no vote.

So far I have this query:
SELECT DISTINCT d.ID, d.name, d.date, v.typeID AS vID, t.name FROM decisions d, votetype t LEFT OUTER JOIN votes v ON v.decisionID = d.ID WHERE (v.personID=1 OR v.personID IS NULL) AND (v.typeID=t.ID OR v.typeID IS NULL) ORDER BY d.date;

For some reason the DISTINCT doesn't work and I get a decision that is not voted for three times (once for each type while v.typeID column is NULL).

How can I fix this so each decision only shows up once?

Jurgen

-------------------------
I've found a solution:

SELECT DISTINCT d.ID, d.name, d.date, v.typeID AS vID, IF(v.typeID IS NULL,NULL,t.name) AS stem FROM decision d, votetype t LEFT OUTER JOIN votes v ON v.decisionID = d.ID WHERE (v.personID=1 OR v.personID IS NULL) AND (v.typeID=t.ID OR v.typeID IS NULL) ORDER BY d.date;

A Simple Outer Join
Sorry to post a naive question , But i need a query for this simple condition.
There are 2 tables employee and email . both have employee id as PK. I need to list employee names and emails . If enrty does not exist in email ..then it should get NULL .
Select employee.name, email.id from employee , email where ?

Left Outer Join
^ just a quick question I can't find an answer for, with the mysql syntax is 'left outer join' the same as 'left join' ?

LEFT OUTER JOIN
am using mysql and want a recordset to be displayed from a query that links three tables.
because a user can add to their selection at a later date i would like all records from a tbl_fitopt to be displayed even when there are no foreign keys in the other tables.
sql is below but at the moment it only lists records from tbl_fitopt that have keys in tbl_fitoptlink - nb i need all records from tbl_fitopt to display so a user can add/subtract from there choices.

strsql = "SELECT tbl_fitopt.fitoptid, tbl_fitopt.fitoptcost, tbl_fitopt.fitopt, tbl_stock.stockid, tbl_fitoptlink.fitoptlinkid "_

& "FROM (tbl_fitopt INNER JOIN tbl_fitoptlink ON tbl_fitopt.fitoptid = tbl_fitoptlink.fitoptid)
INNER JOIN tbl_stock ON tbl_fitoptlink.stockid = tbl_stock.stockid "_

& "WHERE tbl_stock.stockid = " & tkey & ";"

OUTER JOIN With Extra Conditions?
I need to get a list of products in a certain category, along with the quantity of each item already added to the shopping cart for a given session number. I'm having a heck of a time satisfying the latter condition.

shoppingCart table:
+--------+-----------+-----------+-----+
| cartID | sessionID | productID | qty |
+--------+-----------+-----------+-----+
product table:
+-----------+--------------+------+--------+
| ProductID | ProductCatID | name | Active |
+-----------+--------------+------+--------+
Here's my base query, which just gets all the products in a given category.

SELECT product.*
FROM product
WHERE `ProductCatID`='{$id}'
AND `Active`=Ƈ'
Here's the query I've got so far:

SELECT product.*, shoppingCart.qty
FROM product
LEFT OUTER JOIN shoppingCart ON shoppingCart.productID = product.ProductID
WHERE `ProductCatID` = Ɗ'
AND shoppingCart.sessionID = '{$sessionID}'
AND `Active` = Ƈ'
Obviously, this does NOT work, because it limits the query to ONLY products that have the specified sessionID. I need it to return ALL the products in the category, but give me the quanity for items in the shoppingCart table, ONLY IF the sessionID matches (otherwise it should return NULL)!

Multiple LEFT OUTER JOIN
What is the right way to do a LEFT OUTER JOIN on both these tables to the 'Main' table where WHERE Plan.pID IS NULL and Record.rID IS NULL (primary's)

Plan
pID | _mn (foreign main key) | pType

Record
rID | _mn (foreign main key) | rDate

Main
mID (primary) | mComp

-------------------------------------
Will need: pType, rDate, mComp values as a result of the query.

Struggling With Left Outer Join
got two tables:

T1 - User, Date, Stuff

T2 - User, Date, Other Stuff

I want to be able to select rows from T1 based on User & Date, join those rows to the data in T2, so T1.User=T2.User AND T1.Date=T2.Date. Then I want to sort the results in T1.Date order

I've no idea where to start ... anyone point me in the right direction?

Transform SubSelect In OUTER JOIN
maybe I'm simply to dump but I could not transform this SQL-Statment
which uses a Sub-select and create on that uses an OUTER JOIN ....

Double OUTER JOIN On The Same Table?
I'm trying to make a request with an OUTER JOIN on 2 fields at the same time:

Here are the tables:

test:
- id
- user1
- user2

users:
- id
- username

And Id like to get as result each line of the 'test' table, with the corresponding username from 'user' fir the fields user1 and user2.

Do you know how to do this without using several SELECT?

Performance Left Outer Join
I changed a query from an equal join to a left outer join but then the performance is suddenly 10 times as slow ?? Is this normal or do I miss something ??

$query1="SELECT * FROM ";
    $query1=$query1 . "clienttable,contacttable AS co1 WHERE co1.coclid=clid ";
    $query1=$query1 . " AND clname like '$alfabetisch%' ";
    $query1=$query1 . " order by $o";

Full Outer Join Should Work
why doesn't the following work?

mysql> select * from girls full outer join boys where girls.city = boys.city;

where :

drop table girls;
drop table boys;
create table girls (name varchar(12), city varchar(12));
create table boys (name varchar(12), city varchar(12));
insert into girls values('Mary', 'Boston');
insert into girls values('Nancy', null);
insert into girls values('Susan', 'Chicago');
insert into girls values('Betty', 'Chicago');
insert into girls values('Anne', 'Denver');
insert into boys values('John', 'Boston');
insert into boys values('Henry', 'Boston');
insert into boys values('George', null);
insert into boys values('Sam', 'Chicago');
insert into boys values('James', 'Dallas');

If needed, I can go into insert and create for you!

BTW left and right work OK!

BTW what do P1(T1,T2), P1(T1,T2) and R( mean in the following

SELECT * T1 LEFT JOIN T2 ON P1(T1,T2)
WHERE P(T1,T2) AND R(T2)

Left Outer Join Produces
In order to eliminate duplicate values from the right of a query I tried the following:

LEFT OUTER JOIN
        `private_comments`
    ON
        `report_cards`.id IN (select distinct `private_comments`.student_id1 from `private_comments`)

I have unique `report_cards`.id but multiples of `private_comments`.student_id1.

Sould Someone Verify LEFT OUTER JOIN?
could someone please verify :

SELECT *
FROM Members, Submits, MNotify LEFT OUTER JOIN MemPay as mp on
(mp.MPMem_ID=Members.MemID and Members.MemID=colname and mp.MPSub_ID=MNotify.FK_SubmID)
WHERE Submits.SubmID=MNotify.FK_SubmID and Members.MemID=colname
MemPay - payment table, where
MPID | MPMem_ID (actually foreign) | MPSub_ID

Members
MemID

Submits - submitted ads
SubID

MNotify - Members become notified of individual submitted ads
MNID | FK_MemID | FK_SubmID

Members can view each MNotify.FK_SubmID (submitted ads -- the ads may be bought, and if so are found in MemPay table. MPMem_ID is the buyer primary id and MPSub_ID is the id of the submitted ad bought.)

In any case, the Member can view all submitted ads (including purchases) while the check on the LEFT OUTER JOIN values gives what he or she has bought.

While it seems to output as it should, when I check on the LEFT OUTER JOIN values with php conditional statements I've getting values oppositie of what they should (checking on NULL, <>NULL, empty, !empty)

Workaround For Outer Join Using Filtering Subquery
I'm using mysql 4.0.x, which does not allow the following query (which works
fine in 4.1). Is there any way I can issue a single query to achieve the
same results? (I want a list of all records from table al, nulled where
there is no match in table alm, which has been filtered. Without a
subquery, the filtering occurs after the outer join, and for one specific
row from alm, I only see the nonmatching records from al where NO other alm
record matches.)

SELECT al.id, alm.idmember, al.listname
FROM addresslists AS al
LEFT OUTER JOIN
(SELECT idlist, idmember FROM addresslistmembers WHERE idmember = 4)
AS alm
ON al.id = alm.idlist

Left Outer Join Doing Wired Things
i have two tables, tableA and tableB, i want all of tableA but regardless of if there is a record in tableB that references the record of tableA, so thats what an outer (or left, or left outer) join should do if tableA is the left table... right?

Ok first my Query:

SELECT *
FROM tableA A
LEFT OUTER JOIN tableB B
ON A.ID=B._IDa
WHERE ((B._IDb IS NULL) OR (B._IDb=3)); -- no record exists in tableB with _IDb=3

tableA is referenced by tableB's foreign key field _IDa
tableA has 27 rows, tableB has many rows but none that have _IDb=3

So I expect to get 27 rows with NULL for all fields of tableB, but i only get 18 rows!!
(all fields of tableB are, as expected, NULL, but WHY those 18?!! why not all or none?!)

Note though that B is a composite entity, where _IDa is only half of the primary key, _IDb is the other half referencing someother table that's not important for this problem.

Code:

Slow Execution For A Left Outer Join Query
Whats likely to be the cause of slow execution for a left outer join query?

The original query joins three tables but even if I narrow it down to one it still takes a long time to execute.

$query = "select distinct materials.* from materials";
$query .= " left outer join materials_products on materials.material_id = materials_products.material_id";

There's 914 rows in the materials table and 1348 row in the materials_products table

Is it likely to take a long time for this amount of data or is there likely to be a problem in the table(s) set up or query?

You Have An Error In Your SQL Syntax; Check The Manual That Corresponds To Your MySQL Server Version For The Right Syntax To Use Near 'orange' WHERE `ID` = '11128'' At Line 1
I am getting this error message everytime I run my php script....
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'orange' WHERE `ID` = '11128'' at line 1

It seems that there is a limit on the where syntax. I don't know how to fix it, and it's a very big problem for me.

My task is to insert 1million values into mysql table with a php foreach loop. If I insert aprox. 10.000 values, this error don't come, but any numer over will give my mysql error.

Create Table Error. #1064 - You Have An Error In Your SQL Syntax; Check The Manual That Corresponds
I have been trying to create two tables with the SQL below. I have the SQL in file and tried to import it to PhpMysqlAdmin in my control panel. The wierd thing is that the first table gets created while the second one is not created and error is thrown instead. The error message is

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ....

Join Syntax
I'm having a terrible time trying to figure out the proper syntax for this command. I have two tables I want to use:

The first (Player) has the following fields: Name, Season, Game, TeamNumber, Pts, Reb
The second (Teams) has the following fields: TeamNumber, Season, Game, WL

Not all teams have played the same number of games. I would like to be able to come up with a statement that enables me to sort the teams by average points per game (i.e. the average of the total of the players on that team, per game). Is there a way to do this in MySQL 4.0.27? My host won't upgrade so I can't use subselects, and I feel a little trapped here.

I'll be displaying the output via php, and I'd like to be able to be able to sort by Team Wins, Team Losses, Team Avg PPG, Team Avg RPG, and so on. Can anyone give me a bit of help here?

If I use SUM, I end up getting all of the values, so a team that has played 5 games with a low ppg average still ends up higher than a team that has played 1 game and scored a ton. If I use average, I get the averages for each player, not the averages for the team. Any suggestions?

JOIN Syntax
I've read the manual, but it's vague on what the different types of joins actually mean, in terms of concept
How is LEFT JOIN different from RIGHT or OUTER JOIN?

How is JOIN different from a SELECT WHERE clause that explicitly indicates a common field?

Join Syntax
this type of join syntax is giong to work. I've never used a table mulitple times in a join and don't know if you can even do that. I'm in a situation where i can't really test it but it needs to be correct.PHP Code:

select incidents.ref_no, a.label as status_label, b.label as product_label, c.label as sub_product_label from labels as a, lables as b, labels as c, incidents left join labels on (incidents.status=a.label_id) left join labels on (incidents.product=b.label_id) left join on (incidents.sub_product=c.label_id)

Mysql Syntax+As+JOIN
fotos is a table inside a database
##########################

"SELECT *

FROM fotos AS f1

JOIN fotos AS f2

ON f2.id_evento = f1.id_evento

AND f2.arquivo = f1.arquivo

AND f2.id_foto > f1.id_foto

";

Is the above mysql command syntax correct? seems to not work.

MySQL Error, #1064 - You Have An Error In Your SQL Syntax.
I was adding a table in phpMyAdmin and when i clicked on go it produced this SQL CREATE

TABLE `pm` (
`to` VARCHAR( 30 ) NOT NULL ,
`from` VARCHAR( 30 ) NOT NULL ,
`message` BLOB( 25000 ) NOT NULL ,
UNIQUE (
`to` ,
`from`
)
)
and this error
MySQL said:

#1064 - You have an error in your SQL syntax.**Check the manual that corresponds to your MySQL server version for the right syntax to use near '(25000) NOT NULL, UNIQUE (`to`, `from`))' at line 1

Join Syntax And Field Overrides
I am looking for a way to override the values of a record with those pulled in from a join. However, if the fields of the override are NULL, I want the original value to stay intact. Is this possible?

Example:
SELECT t1.*,
if(t2.item_title is null, t1.item_title, t2.item_title) as item_title,
if(t2.item_description is null, t1.item_description, t2.item_description) as item_description,
if(t2.item_summary is null, t1.item_summary, t2.item_summary) as item_summary
FROM catalog_item AS t1
LEFT JOIN catalog_item_text AS t2
ON t1.item_id = t2.item_id
AND t2.item_langid = 'en_us'
WHERE t1.item_id = 1

In the above example, a record keyed on a Language ID (for localized text) is patched in through a join. If a record is found, it contains three text fields that MAY contain a title, description, or summary for that item.

The problem is if a particular field is NULL I want the original default value to be used instead. The above example does solve this problem, but what i am looking for is a better way where I do not have to explicitly know the fields involved as long as they exist in BOTH tables.

I have been banging my head against the wall on this for a while now, finally decided to see if anyone here might have an answer. Any thoughts?

Java Code Or Pseudo Code For "Outer Join"
I want to implement the "outer join" functionality in Java. Can
somebody explain the pseudo code for the same. OR what needs to be done
to extend the hash-join Java code of equijoin.

JOIN Syntax (INNER JOIN), Self Join
I want to find the most recent purchase for each customer(see below) and I tried:

SELECT A.*
FROM mytable A INNER JOIN (
SELECT full_name, Max(entered_when) AS entered_when FROM mytable GROUP BY full_name
) M ON A.full_name=M.full_name AND A.entered_when=M.entered_when
;

I got a syntax error:
ERROR 1064: You have an error in your SQL syntax.  Check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT full_name, Max(entered_when) AS entered_when FROM mytable

My data looks like:
full_name  product_purchased  entered_when
phil          shoes             20050227121012
mary          purse             20020212000000
phil          socks             20021226101423
mary          bra               20020111000000

(I just want to get phil's shoes and mary's purse.)

What should I be doing?

ERROR 1064: You Have An Error In Your SQL Syntax Near
Now this is odd. I'm running mysql Ver 12.22 Distrib 4.0.18, for apple-darwin6.8 (powerpc) right?

And I can't get any subquery to work. I even tried :

SELECT (SELECT 1);
and I still get

ERROR 1064: You have an error in your SQL syntax near 'SELECT 1)' at line 1

ERROR 1064: You Have An Error In Your SQL Syntax
I am getting error while executing below Query. I have these two tables and working in individual queries. I need to combine and get the the data.

SELECT DLVRY_PORT FROM TENDER WHERE DLVRY_PORT IN
( SELECT PORTID FROM SUPPLIER_PORT WHERE userid = 'sammil' )

Error:

[root@localhost:3306] ERROR 1064: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT PORTID FROM supplier_port WHERE userid = 'sammil')'

"ERROR 1064 (42000): You Have An Error In Your SQL Syntax " When Trying To Import Backup
Just upgraded to the latest version of mySQL. Before I did that, I did a full backup of all tables using the Administrator tool. Now when I try to do a restore, I get the message in the subject. Even if I copy-paste the SQL code into the command line client, I get the error. For example, I use this code:

CREATE TABLE "news" (
"news_id" int(4) NOT NULL auto_increment,
"news_subject" varchar(96) NOT NULL default '',
"news_author" varchar(96) NOT NULL default '',
"news_email" varchar(96) NOT NULL default '',
"news_body" longtext NOT NULL,
"news_time" int(10) NOT NULL default '0',
PRIMARY KEY ("news_id")
) TYPE=MyISAM DEFAULT CHARSET=latin1;

And I get this error:

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use near '"news
" (
"news_id" int(4) NOT NULL auto_increment,
"news_subject" varchar(96' at line 1

Syntax Error
What is wrong with this query?:

CREATE TABLE `cms_config` (
`site_name` VARCHAR( 25 ) UNSIGNED NOT NULL ,
`site_url` VARCHAR( 50 ) UNSIGNED NOT NULL ,
`site_id` TINYINT UNSIGNED NOT NULL AUTO_INCREMENT ,
`site_desc` VARCHAR( 64 ) UNSIGNED NOT NULL ,
PRIMARY KEY ( `site_id` )
) TYPE = MYISAM
That is the SQL phpMyAdmin has generated and it is giving me this error message:

Quote:

#1064 - You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'UNSIGNED NOT NULL ,
`site_url` VARCHAR( 50 ) UNSIGNED NOT NULL

I checked the MySQL manual and all it tells me is that it's a syntax error

Syntax Error
I was trying to create a table in my database, but it kept coming up with syntax errors, and that I should check the manual.
I have used many different data types including varchar, char, text and int. But I keep getting syntax errors.
Please can someone help and tell me what data types I should be using for mySQL version 5.0.

Syntax Error
~ RESOLVED ~ Can't figure out where the error is.

query:

$sql="INSERT INTO response(`consId`, `critique`,`comment`,`provId`) VALUES ('$consId','$critique','$comment','$provId') WHERE provider.provname='{$_POST['provider']}' AND consumer.name='{$_POST['sname']}'";
returns error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE provider.provname='Healthnet' AND consumer.name='adfadfasd

SQL Syntax Error???
The code below is suppose to display the month and year of the articles in groups. At the moment I am getting the following when I test the
Could not query because: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1.
Can someone please have a look at my code and tell me why this is happening?

PHP

require "config.php";
// display the archives
$query2 = mysql_query ("SELECT * FROM tbl_articles ORDER BY STR_TO_DATE( CONCAT( day, month, year ) , '%d%M%Y'") or die ("Could not query because: ".mysql_error());
while($row2 = mysql_fetch_array($query2)){
echo "<a href=archives.php?month=".$row2['month']."&year=".$row2['year'].">".$row2['month']." ".$row2['year']."</a>";
}
mysql_close();

Syntax Error?
I get this error when this code is run

PHP

$sql = "UPDATE PROVIDER SET Approve = Ƈ',IP_Address = '$IP',Name = '$Name',Email = '$Email',Website = '$Website',Phone = '$Phone',Display_Email = '$Display_Email',Website_URL = '$Website_URL',Display_Phone = '$Display_Phone',Age ='$Age',Eye = '$Eye',Rate = '$Hourly_Rate',Location = '$Location',Other_Location = '$Other_Location',Hair_Color = '$Hair_Color',Build = '$Build',Hair_Length = '$Hair_Length',Bust = '$Bust',Hips = '$Hips',Cup = '$Cup',Waist = '$Waist',Height = '$Height',Availability = '$Availability',Grooming = '$Grooming',Services = '$Services_Offered',Rules = '$Rules',Who = '$Who',From = '$From',To = '$To'WHERE ID = '$ID'";echo $sql;

And heres the output
UPDATE PROVIDER SET Approve = Ƈ', IP_Address = པ.161.187.25', Name = 'llllllllllllllllllll', Email = 'test@aol.com', Website = 'No', Phone = '', Display_Email = 'No', Website_URL = '', Display_Phone = '', Age ='', Eye = '', Rate = '', Location = 'Las Vegas, NV', Other_Location = '', Hair_Color = '', Build = '', Hair_Length = '', Bust = '', Hips = '', Cup = 'N/A', Waist = '', Height = '', Availability = 'In-call', Grooming = 'Clean Shaven', Services = 'N/A', Rules = '', Who = '<br>', From = &#55614;&#57159;-09-12', To = &#55614;&#57159;-09-12' WHERE ID = ྞ'You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'From = &#55614;&#57159;-09-12', To = &#55614;&#57159;-09-12' WHERE ID = ྞ'' at line 29

SQL Syntax Error
I'm getting the following error, but only for certain queries:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Sometimes the query works fine, sometimes it doesn't. I've tracked a few and looked at which record they hung on, but nothing obvious.

I have the exact same script running on two different MySQL DB's. One works, one doesn't, sometimes. What could the difference be in what I thought was exactly similar DB's?

SQL Syntax Error
I got an error "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1" on
$record = mysql_query($sql);
if(!$record)
{
die('no record!:' .mysql_error());

I cannot figure out what is wrong with this statement. I am new to mysql and I am trying to retrieve key and value from a table