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Cannot Get MySQL Query To Work In .asp


I have a MySQL database and in the front-end software I am using to Run MySQL, I have created a query on the tables which runs in the front-end, similar to the MS Access query grid.

I now wish to pull this query on-line using .asp, but I seem to have found out that I can only pull data from tables in MySQL? ....




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Related Forum Messages:
Degrading Sub Select Query To Work In MySQL 4
I have the following query working as intended in MySQL 5. However, I had not realised that the target server is running 4.0.2 and it's not possible to upgrade.

SELECT
records.player_id,
records.record_id,
records.game_id,
records.updated,
game_modes.mode_title,
records.record_time,
players.firstname,
players.lastname,
players.nickname,
players.country,
teams.team_title,
classes.class_title,
tracks.track_title,
games.game_title,
games.version,
record_types.type_title
FROM games, records, classes, teams, tracks, record_types, players, game_modes
WHERE
records.record_time = (
SELECT MIN(record_time)
FROM records
WHERE records.player_id = players.player_id
AND records.game_id = 1
AND records.mode_id = 1
AND records.track_id = 1
AND records.class_id = 5
AND records.type_id = 1
GROUP BY records.player_id)
AND

records.track_id=tracks.track_id
AND record...................

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What Query Would Work For This?
I have two tables: say Yes and No.

Yes(ID, YesEmail, EmailOwner);
No(ID, NoEmail);

Both YesEmail and NoEmail contain email addresses.

What query can retrieve all the email addresses in Yes table that are not listed in No table? Thanks in advance.

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Query Wont Work
PHP Code:

if(mysql_query("UPDATE users SET referer_score=referer_score+1 WHERE userID='$frommysql_referer'"))
{
               die(mysql_affected_rows());

im trying to figure out why refer_score isnt being iincremented as im expecting it to. i know that the value of $frommysql_referer is the right value that im expecting it to be. just it wont increment referer_score. and mysql_affected_rows() wont print anythnig to the screen

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Query Doesn't Work With Mysql4
Can someone tell me what's wrong with the mysql query below? It does work on mysql 5 but gives me an error with mysql 4.

SELECT rev_profile_indiv. *
FROM rev_profile_indiv, rev_profile_kids_indiv
WHERE 1
AND reviewing
IN ( 1, 0 )
AND active
IN ( 1 )
AND contract
IN ( 1 )
AND UNIX_TIMESTAMP( DATE( rev_profile_kids_indiv.birthdate ) ) >= UNIX_TIMESTAMP( ��-12-15' )
AND UNIX_TIMESTAMP( DATE( rev_profile_kids_indiv.birthdate ) ) <= UNIX_TIMESTAMP( &#55614;&#57158;-12-15' )
GROUP BY rev_profile_indiv.username
ORDER BY lastname
LIMIT 0 , 30;

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What Do I Need To Do To Make This Query Work?
I must've been working too long hours this week, as my flow of ideas has come to a complete stop.

Its really quite simple: 2 tables - bookings , invoices

These two counts gives me precisely what I want

PHP

$unpaid_bookings = mysql_result(mysql_query("SELECT COUNT(id) FROM bookings    WHERE invoice_id = 0 AND client_id = $id AND status != &#390;' AND finish <= '$time'"),0);$total_bookings = mysql_result(mysql_query("SELECT COUNT(id) FROM bookings    WHERE client_id = $id AND status != &#390;' AND finish <= '$time'"),0);


so I present this in a ratio, $unpaid_bookings / $total_bookings... awesome

now, when I try and flip the coin and show unpaid invoices over total invoices, it all falls apart once I have more than 1 booking per invoice.

First i'll explain the table structure. Each booking has a column `invoice_id`, which is assigned to the primary `id` of the invoice table. So one invoice might be associated with x bookings.

PHP

$unpaid_invoices = @mysql_result(mysql_query("SELECT COUNT(i.id) FROM invoices i, bookings b     WHERE i.id = b.invoice_id AND b.client_id = $id AND i.is_paid = &#390;'"),0);$total_invoices = @mysql_result(mysql_query("SELECT COUNT(i.id) FROM invoices i, bookings b     WHERE i.id = b.invoice_id AND b.client_id = $id"),0);


Again so it works fine when 1 invoice is associated to 1 booking, but when 1 invoice is associated with say 2 bookings, the invoice count will return two instead of one. The worst part is, I know exactly why (because there are two bookings for the invoice), I just can't get around it!

I've tried all sorts of joins, group by's, and i'm not having any luck. I'm trying not to be weak and add the client_id into the invoice table,

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Query Does Not Work On Some Server Instances
query:

Select * From Students Where C_Name = "XXXXXX"

WHen I run this query on 2 of my computers, the result returns no rows.

When I run this query on a remote machine hosted by 1and1, with a copy of the same database, the query returns the correct row.

The syntax is identical.....

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Cant Work Out Correct Select Query For ..
mysql> select * from parent_cat;
+-------+----------+----------+
| catid | pname | psection |
+-------+----------+----------+
| 1 | Trainers | Men |
| 2 | Bags | Woman |
| 4 | Boots | Woman |
| 5 | Fragance | Men |
+-------+----------+----------+
4 rows in set (0.00 sec)

mysql> select * from categories;;
+-------+---------------------------+---------+------+
| catid | catname | section | pid |
+-------+---------------------------+---------+------+
| 1 | Nile | Men | 1 |
| 2 | Bags | Woman | 2 |
| 4 | Boots | Woman | 4 |
| 6 | Air Force I & II Trainers | Men | 1 |
| 7 | Adidas Trainers | Men | 1 |
| 8 | Timberland Trainers | Men | 1 |
| 9 | Lacoste Trainers | Men | 1 |
| 10 | Gucci & Hogen Trainers | Men | 1 |
| 11 | D&G Trainers | Men | 1 |
| 12 | Prada Trainers | Men | 1 |
| 13 | Chanel Trainers | Men | 1 |
+-------+---------------------------+---------+------+

and the following query:

$pid = $_GET['catid'];

$query = "SELECT categories.catid, categories.catname, categories.pid
FROM categories, parent_cat
WHERE categories.section = 'Men'
AND parent_cat.catid=categories.pid
AND categories.pid = '$pid'";

At the moment the query only returns all categories.catname when my categories.catid = 1, but returns nothing if it equals something else?

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Query Which Gets Rows Based On A Radius, Lon And Lat Doesn't Work?
I have found this query which gets rows based on a radius. I need this for zip codes based on lon and lat's.

$sql2 = "SELECT * FROM table as z WHERE (SQRT( (69.1 * (".$userLat." - z.lat)) * (69.1 * (".$userLat." - z.lat)) + (53.0 *(".$userLong." - z.lon)) * (53.0 *(".$userLong." - z.lon))) <= ".$userRadius." )";
return $sql2;
$res = mysql_query($sql2, $connDB) or die(mysql_error());
$row = mysql_fetch_assoc($res);
based on a test zip code gives a result like


SELECT * FROM table as z WHERE (SQRT( (69.1 * (52.399834 - z.lat)) * (69.1 * (52.399834 - z.lat)) + (53.0 *(4.840762 - z.lon)) * (53.0 *(4.840762 - z.lon))) <= 100 )
the lat and lon of the test zip code are right. As you can see z.lat and z.lon don't get any value. And these would be every lat and lon in table.

In my db table lon and lat are decimal(10,6) type with a default value of 0.000000

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Aggregated Variable Only Work With Transaction In Query Browser?
I want to select rows and use user variable to generate an incremental id for the result. For example, I want the following query

select @id:=(@id+1) as id from file_list limit 5

to return 5 rows with values 1 to 5. However, the result is ....

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Reformatting Email Address To Work On SELECT Query
This is probably basic, but so far I cant find any articles about this.

I want to "SELECT * FROM dbase WHERE emailaddress = $email;"

$email is a feed from PHP, and will contain a string like "some.one@someplace.com"

Apparently with $email as a normal string, it gives an error saying:
"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 '@someplace.com)' at line 1"

I reckon it's the dots and the @s that terminate the syntax and making it not work.

How can i get out of this problem? Is there any function I can call to solve this? Or probably a little bit of more programming?

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Work In MySql
This query works fine in SQL Server, will it work in MySql?

SELECT a.total, b.browsertotal, b.browser, (b.browsertotal / a.total) * 100 AS browserpercent
FROM
(SELECT Count(id) As total
FROM MyTable) a,
(SELECT count(id) as browsertotal, browser
FROM MyTable
GROUP BY browser) b

I dont have MySQL and I am trying to help someone out.

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Getting PHP &amp; MySQL To Work Together
I am trying to get PHP to write into MySQL database. Right now it does not seem that it is doing such, MySQL doesnt appear when I bring up the PHPInfo() page. I have removed semi-colons before the Mysql.dll extension in php.ini, and I have made sure that PHP is in the windows PATH. What else do I need to do to get this stuff running?

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Getting MySQL To Work..
I downloaded version 4.1.7, opened the manual, followed the instructions, and installed it. Then, when I opened up the "MySQL Server Instance Config Wizard", I set it all up. I press the "Configure" button, and it runs up the point where it creates the Service. Then, it gives this message"Could not start the service MySQL. Error: 0".

I've ran a Google search on the error message, but have found nothing.

I'm running Windows XP. If there's any more information you need about my computer setup, go ahead and ask.

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Mysql 5 Help Doesn't Work
when i type "help show;" in mysql 5, nothing shows up. Why can i not see it.?

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MySQL Doesn't Work In XAMPP
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MySQL To Work On My Localhost
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MySQL Join Does Not Work
please take a look a this query:

select * from predmeti p, rjesenja r WHERE p.predmetID = r.predmeti_predmetID

As you can see, here I want to display content from table predmeti and rjesenja. But only if there is a foreign key in rjesenja.

This works fine, it shows the desired results, but..

the vise versa version:

select * from predmeti p, rjesenja r WHERE p.predmetID != r.predmeti_predmetID

does not work, here I want to display all from table predmeti, but only if there is
no foreign key in table rjesenja.

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FIND_IN_SET Doesn't Work In MySql 4.1.9
I need some help with FIND_IN_SET function. I'm not able to use this function in MySql 4.1.9 but in MySql 4.0.x I can !!! I've installed the latest version of EasyPhp.

When I use a SQL request with FIND_IN_SET it doesn't succed!!

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How Many Table Can MySql Work With
I would like to generate 1,000,000 tables. But can MySql work with that many tables?

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MySQL Tabs Work
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What are the main ones, and what am I supposed to do in the main ones?

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Mysql 4.1 Inner Select Dosen't Work
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--------------------------------
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--------------------------------

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MySQL Admin Backups Won't Work
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That sounds straight forward enough – it works like every other gui out there. The problem is you click on the database you want to backup and it doesn’t become highlighted and when you click on the”>” button, nothing happens. Why?

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Mysql Server Wont Work
i am having mysql install with Plesk 7.5 for windows
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I have the Fedora version from the Wiley book on Linux. I am having problems with getting phpmyadmin to work with mysql. I get "No Privileges" when I try to create a database.

All I see is a test database. I don't see a "mysql" database. I think something did not get installed correctly in the initial setup.

Every time I try to grant privileges, change passwords, I get access denied.

I am logged in as root.

Can someone provide instructions on initial setup of mysql or point me to a script I can run that will set up the mysql database for me?

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I've set up a DSN for the database, under ODBC Data Sources, and I'm using a virtual directory under IIS for the web pages I'm developing. When setting up a database connection though in Dreamweaver, I get an error message that says "An unidentified error has occurred" As far as I can tell, I have MySQL set up for Root access and also anonymous access. Any suggestions about what I'm missing to get this to work?

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Reg MySQL Server Work As Local Network Or Remote Server
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Getting Dreamweaver 8 To Work With MySQL: The "Unidentified Error"
I set up MySQL on my Macintosh at my job (G4 / OSX 10.3.9) 3 days ago, along with PHP --

all went fine, & I got MySQL started, a DB set up -- until I tried to connect Dreamweaver to it I go to connect to the DB in Dreamweaver 8, and when I try to set up the DB in the "databases" tab and click on the "database" field I get :

"An unidentified error has occurred" Code:

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Mysql 4 Vs Mysql 5? Why Does This Not Work In Both?
$search = "SELECT ct_vehicles.id,".
" ct_vehicles.description,".
" ct_vehicles.price,".
" ct_vehicles.manufacturer,".
" ct_vehicles.new,".
" ct_sites.county,".
" ct_images.id,".
" ct_images.vehicleId".
" FROM ct_vehicles".
" LEFT JOIN ct_sites".
" ON ct_vehicles.siteId = ct_sites.id".
" RIGHT JOIN ct_images".
" ON ct_vehicles.id = ct_images.vehicleId".
" WHERE (".$where.")"
" LIMIT ".$offset.
",".$rowsPerPage.";";

$where = my condition set (this is correct i think)
$offset = Offset set by my paging alongside Rows per Page
$rowsPerPage = As above;

In mysql 5 it displays the correct 35 records where as in mysql 4 it displays some 13000 records WTF.

SELECT ct_vehicles.id, ct_vehicles.description, ct_vehicles.price, ct_vehicles.manufacturer, ct_vehicles.new, ct_sites.county, ct_images.id, ct_images.vehicleId FROM ct_vehicles LEFT JOIN ct_sites ON ct_vehicles.siteId = ct_sites.id RIGHT JOIN ct_images ON ct_vehicles.id = ct_images.vehicleId WHERE (ct_vehicles.description LIKE '%red%' OR ct_vehicles.manufacturer LIKE '%red%' OR ct_vehicles.colour LIKE '%red%' OR ct_vehicles.text LIKE '%red%' ) LIMIT 0,5;

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Bug Work
I've got this bug http://bugs.mysql.com/bug.php?id=27861. I know they've made a fix and it's going to come out in the next version but is there any way I can fix this myself?

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How Does Sql Work Exactly?
Well what exactly does sql do when you create a table for like a login (user) What does the tables and stuff do. Do they just create like a library to store that users information so they can log back in to site or w/e....

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Subselect Does Not Work
i am trying to remove values from a list menu if the join table d=
oesnt have keys when a key is selected for instance:
locations
locationID

locations_join
locationID shotlistID

SELECT SQL_NO_CACHE l.locationID , l.location FROM locations l LEFT JOIN lo=
cations_join lj ON l.locationID =3D lj.locationID WHERE l.locationID NOT IN=
(select locationID FROM locations_join WHERE shotlistID IN (5069)) ORDER B=
Y l.location ASC.
so when shotlistID is selected all the keys from the locations_join joined =
to the shotlistID would be remove from the locations list please help, i'm =
trying to do this in one query saving from getting all the keys into an arr=
ay then checking if the values arent in the array when generating the list.

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Date_add Does Not Work
I'm using mySQL 5.0, with a database of UNIX timestamps. Can someone explain this to me? This query works great:

 SELECT from_unixtime( 1132277574 ) 

But this one breaks:

 SELECT date_add(from_unixtime(  1132277574 ), interval 1 week)

Telling me, oh-so-helpfully, that the error is near "'week)'". (It doesn't work when I use "1 weeks" either.)Does the from_unixtime simply not work with the date_add function, or is there some arcanity to this syntax that I'm missing?

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I am putting
log = "C:Program FilesMySQLMySQL Server 5.0sql.txt"
into my .ini files and mysql wont start.
I cant work out the syntax to tell mysql to ouput all the queries to a log file.

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How Does Synchronizing Work?
I have a question;

currently we want to run our central database from two different locations as opposed to one currently. I am trying to figure out how to synchronize the two locations.

My question is: if two (or more) locations create new clients locally, each location will thus create a new client-numbers on their own active operational database. This opens the possibility that each locations will create identical client-numbers; of course being different clients altogether.

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Mysql_connect Does Not Work
I have searched through the forum and tried the suggestions that have been made for this issue but I am about ready to give up! I have installed Apache 2.0.52, PHP 5.0.3, and MySQL 4.0.23 on Windows XP. I uncommented extension=php_mysql.dll in the php.ini file and copied libmysql.dll into both C:WindowsSystem32 and C:WindowsSystem.
My phpinfo.php test works fine so I know PHP and Apache are working together. I do not see any MySQL references on that entire page. From previous posts I know I should. Needless to say, mysql_connect(...) does not work, it gives an undefined function error. I apologize if this should have been posted in the PHP forum instead. If it is a PHP issue, I will repost this there.

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Sql Statement Doesn't Work
When i was using a previous version of mysql (3.something) this sql statement worked fine.

strSQL = "SELECT COUNT(*) as COUNTER FROM prayerrequest where dateexpires <= '" & sDate & "'"

sDate equals todays date.

But after installing the new version (4.1) of mysql this statement no longer works.

I went as far as stepping through all dates and comparing them and it wouldn't work. "1/11/2005" from the database did not equal sDate which was "1/11/2005".. The only way I could get it to work was to use the asp/vb function FormatDateTime on dateexpires and then compare it with sDate.

Does mysql not return a different var type? Is there anyway I can convert dateexpires in the above sql statement so I can compare it with sDate?

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Cannot Get 3 Where Statement To Work
I am having a problem with MySQL 5. I cannot get 3 where statements to work with my database. Any two of the three work fine and return records as they are supposed to but when I add a third clause (all clauses are AND) I get absolutely no records at all and no error message. I don't understand what could be causing this and can use any help offered.

SELECT this, that, otherthing FROM myTable WHERE something = "this" AND somethingelse = "that" AND yetsomethingelse > 0 ORDER BY RAND() LIMIT 3;

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Join Doesn't Seem Work
I have some issues with a simple query, but I just can't figure out why the query doesn't return expected results.

Here is the situation:

I have a table look like this:

TRANSACTION(UNIQUE_ID(PK), CUSTOMER_ID, EMP_ID, TRANS_TYPE, TRANS_DATE, POSTED_DATE, AMOUNT_CHARGED, AMOUNT_PAID, TRANS_LOC, TRANS_ID_CHARGE, TRANS_ID_PAYMENT)

Some rules for the data in the table:

- The TRANS_TYPE indicates if it is a charge or payment(C charge, P for payment). For a payment, there is a unique TRANS_ID_PAYMENT, and for same transaction, the charge record holds same TRANS ID in the TRANS_ID_CHARGE field.

- The TRANS_LOC is for location where the charge happened. But for a payment record, this field is null. which means, payment records and charge records can't be linked together through location.

- TRANS_DATE is the date the transaction happened. For a payment record, this field is also null.

- A charge record and a payment record for same transaction usually have different POSTED_DATE since they are posted most likely on different days.

- AMOUNT_PAID could be less than AMOUNT_CHARGED for same transaction since customers can have an installment payment plan.

Here is my query:

SELECT DISTINCT
A.CUSTOMER_ID,
A.AMOUNT_PAID
FROM TRANSACTION A
WHERE A.TRANS_TYPE = 'C'
AND A.TRANS_DATE BETWEEN '2007-01-01' AND '2007-09-30
AND A.LOC_ID = 1
UNION ALL
SELECT DISTINCT
A.CUSTOMER_ID,
A.AMOUNT_PAID
FROM TRANSACTION A,
TRANSACTION B
WHERE A.TRANS_TYPE = 'P'
AND B.TRANS_TYPE = 'C'
AND B.TRANS_DATE BETWEEN '2007-01-01' AND '2007-09-30
AND B.LOC_ID = 1
AND A.CUSTMER_ID = B.CUSTOMER_ID
AND A.EMP_ID = B.EMP_ID
AND A.TRANS_ID_PAYMENT = B.TRANS_ID_CHARGE

The purpose of this query is to pair all the charge and payment records for the same transaction that happened between a certain period of time. But the results show that there are some dangling payment records that are not linked to any charge records.

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Min Max Does Not Work For Me For Timestamp
I have such a table:

id
timestamp
value
person

regardless or entering this
SELECT id, MIN(timestamp),value FROM table GROUP BY person

or this

SELECT id, MAX(timestamp),value FROM table GROUP BY person

The timestamp value may be properly pulled, yet, the value field is always the same. Meaning, not the value associated wuth the MIN or MAX timestamp value is shown, but the very first for that person. How is that possible?

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Why Does This Grant All Does Not Work?
I've created a new user like this:

create user myuser identified by 'xxxxx';
commit;

Now I'd like to give him all privileges to a certain database:

grant all on db210216204.* to 'myuser'@localhost;

Now I get an error message:

ERROR 1133 (42000): Can't find any matching row in the user table

What am I doing wrong? The database exists:

mysql> show databases
-> ;
+--------------------+
| Database |
+--------------------+
| information_schema |
| baka |
| db210216204 |
| mysql |
| test |
+--------------------+
5 rows in set (0.13 sec)

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LD_PRELOAD Doesn't Work
LD_PRELOAD doesn't seem to work with /usr/sbin/mysqld. It works for other programs, but when it comes to mysqld, I always get the same error saying that the object cannot be preloaded, so ignored. I checked that mysqld is not a setuid binary. Is mysqld doing something funny or did I miss anything?

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Auto_Increment Doesn't Work Any More!
After upgrading from 5.0.11.beta to 5.0.15 on a Windows 32 system, I can't use the auto_increment feature any more; for example, phpMyAdmin gives the following error message: "#1366 - Incorrect integer value: '' for column 'ID' at row 1". Other php clients also (WordPress, MediaWiki...) don't work.

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Logging Doesn't Work
I have WinXP, and MySQL 5.0.18 running as a service. I followed the instructions for
setting up general query log file (mysqld --log), and the file was created (Nebo.log),
but it does not display any logs.
This is all I have after executing a few queries:

Nebo.log

mysqld, Version: 5.0.18-log. started with:
Tcp port: 3306 Unix socket: (null)
Time Id Command Argument

Can anyone offer any help?

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WHERE EXISTS Does Not Work.
SELECT id,name FROM tcesports_groups WHERE EXISTS (SELECT * FROM tcesports_teams WHERE tcesports_teams.gid=tcesports_groups.id);

i get this:

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 'EXISTS (SELECT * FROM tcesports_teams WHERE tcesports_teams.gid

Strucuture of the both tables:
tcesports_teams
-------------------
gid smallint(6)
img varchar(100)
mod mediumint(9)
pos smallint(6)

tcesports_groups
-------------------
id smallint(5)
name varchar(100)

currently the entries for id or gid are just 2digits long (1-99) so im a bit confused what throws this error. im using mysql 4.0 branch

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Difference Of Work
My provider has passed on version 4.1.7 and the inquiry of the following kind has ceased to work for me:

SELECT goods.id, goods.name FROM goods LEFT JOIN category ON goods.id = category.goods AND category.razdel=2 WHERE category.goods IS NULL.

Before all worked correctly, on the version 3.x, what could exchange?
Where a problem?

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Sentence Doesn't Work
I'm using mysql 4.0.21 and Toad to run this query. The message i'm receiving is that there is a Syntax error, where?

select a.id_opcion, a.nombre, a.campo, a.visible, b.id_usuario,
if(b.ancho is null, 50, b.ancho) as ancho, b.posicion
from det_rmen as a
left join
(select *
from rmen_usu
where id_usuario = 'generico') as b
on (a.id_opcion = b.id_opcion and a.campo = b.campo)
where (b.id_usuario='generico') or (b.id_usuario is NULL)
order by id_opcion, posicion;

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How Does 'query_buffer_size' Work In 3.23.56
I see a variable named 'query_buffer_size' that value is 0 in mysql server 3.23.56. I didn't find relational materials in reference. so I wander if any one could tell me what is this variable for?

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