MYSQL :: If Entry Not Found

Mysqladmin Not Found
I have been trying to get mysql installed on a Fedora 3 box for the past week.

I have installed the mysql 4.1 rpm and installed it the system. however I can not use any mysqladmin comands, nor can I shutdown or restart the deamon.

mysqld is running in my processes, and I can connect to it with the url. However I have no idea if its worring correctly, I assume not because I dont have any mysqladmin commands.

No Tables Found
I've recently set up a local test server on my new Mac under PHP5, MySQL 5 (5.0.24-standard), Apache 2, and phpMyAdmin - 2.8.2.2.
When restoring a database from the raw data files from my live remote server's mysql data dir, phpMyAdmin no longer sees any tables in any of my local databases. Instead, I just get the "No tables found in database" message below the database pulldown.
Note that the pulldown correctly lists the number of tables next to the database name (e.g. "myDatabase (5)") in the pulldown. Also note that my local sites work with the updated data, and I can see the tables in MySQL Admin. This seems to be a problem only with phpMyAdmin. Rebooting or turning on register_gobals did not work.

PID FILE NOT FOUND
mysql has been pissing me off today.... ive spent hours trying to solve the problem.. but what I can get it down to is that mysqld daemon fails to start, thus the mysql.sock is not produced where basically the mysql database wont work...the system log when it tries mysql is:

linux mysql: No mysql PID file found. Looked for /var/lib/mysql/linux.server.pid

then it stops mysql

Update If Duplicate Found
Im currently running a statement that inserts a new row into the table if a duplicate row is not found:

PHP

$db->Execute("INSERT INTO tee_time ( SELECT $locationid AS locationid, (SELECT courseid FROM course_conversion WHERE coursename = '$currcourse' LIMIT 1) AS courseid, '$currcourse' AS coursename, '$currtime' AS teetime, '$currdate' AS teedate, '$currcost' AS cost, NOW() AS grabdate FROM tee_time WHERE coursename = '$currcourse' AND teetime = '$currtime' AND teedate = '$currdate' HAVING COUNT(*) = 0)");

The problem is if a row is found I need it to update that row instead of just skipping it. On duplicate wouldnt work because teetime and teedate can be the same in multiple rows.

53 File Not Found: LibmySQL
The server is connected ok but keep getting this error message.

53 file not found: libmySQL.

I have this file in the bin directory: libmySQL.dll.

Mysqld Command Not Found
I followed all the instructions in the tutorial section of the manual, but typing mysql -u root I get Error 2002 Cannot connect thru socket. From what I gather, this is because mysqld is not started (since mysql.sock does in fact exist.) Well, if I try to start mysqld, redhat tells me Command not found. So, I thought I'd try mysqld_safe. Well that gave me some feedback:

Starting mysqld daemon with databases from /var/lib/mysql
STOPPING server from pid file /var/run/mysqld/mysqld.pid
050413 [timestamp] mysqld ended

Update Affecting Only First Found Row
I'd like to write a query along the lines of

UPDATE table
SET field=value
WHERE status=AVAILABLE

Now...let's pretend that there are 4 rows that have status available...how can I just have it affect one row only?

Column Data Not Found
i am having this problem where my html form/php handler are not inserting the correct values into the database. I am trying to simply insert a row into a database using the same php code that i always do. could it be the different mysql structure that i am using? when i click the submit button on the page it inserts a row with all 0s. (-|-|0|0|0.00).

sql structure is as followed. i have never set up my own table with integer fields..

CREATE TABLE `mytable` (
`id` bigint(20) NOT NULL auto_increment,
`date` varchar(100) NOT NULL,
`amt_sold` smallint(5) NOT NULL,
`amt_rem` smallint(5) NOT NULL,
`raised` decimal(8,2) NOT NULL,
KEY `id` (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

No Tables Found In Database
Installed I have:

PHP Version 4.3.10
MySQL 5.0.2-alpha-Max
phpMyAdmin 2.7.0-pl2

Recently upgraded from MYSQL 4.1.

I can loginto phpMyAdmin fine and my databases are listed on the left hand side, however, clicking on any database shows "No tables found in database".

I guess I need to convert all the tables or something but I'm lost with all the MYSQL documentation!

Even creating a new database via shell and checking it in phpMyAdmin I get the same error?!?

Mysqld.sock Not Found
When I enter mysql -u root -p I get this error:PHP Code:

ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (2)

I've searched google and there are hundreds of threads in other forums about this same question. All of them are unanswered that I looked through. I asked in a couple IRC rooms and noone seemed to know exactly what the problem could be. A lot of people were telling me that the socket is probably being created somewhere else but they didn't know how to find out where its being created or how to change it. I checked my.cnf and the sockets are all set to /var/run/mysqld/mysqld.sock.
Does anyone know what the problem is or how to fix it?

Handle No Data Found Situation
I am kind of new and am trying to set up error handlers in MySql
5.0.41.
I create a function. Within, I try to keep selecting until I get
nothing back from select.

I cannot seem to catch the error when the select returns nothing.
I declare a handler like this:
DECLARE CONTINUE HANDLER FOR NOT FOUND
BEGIN
SET tester = 01;
more stufff....
END;

DECLARE EXIT HANDLER FOR SQLWARNING
BEGIN
write out to error log...
END;

Then, I have a loop like this:

myloop: LOOP

select x INTO y
FROM z
WHERE a = b;

if tester = -1 then
LEAVE myloop;
end if;

do stuff here....

set b = b + 1;

END LOOP myloop;

The select is guarenteed to retrieve nothing after a few loops around.
but it never ends. The declare continue handler does not fire.

If I run from client command prompt the EXIT handler fires, but I
dont get the warning message, just message thatit exited without a
RETURN.

If I remove all error handlers and run form prompt, it runs endlessly
until I stop it with a counter var. Then I get X number of warnings,
but it doesnt say what the warning is

Page Can Not Be Found (time Out And Slow_queries)
it seems that im getting a page can not be found a lot when I try to edit the table information. Ive looked at Runtime Information and Slow_queries 5,220 jump out at me is that a reason for this page error? I have a php page that gets its information from the table.

the information changes regularly and also I have the php page delete entries by date every 8 days. I included the code If theres something I can do to make it not load the server up or what ever. I hope all this makes sence. I guess the ? is why do I get page cant be found, when I try to edit stuff in the table? Code:

MAC OS X : 'mysql -u Root' : Command Not Found!
i know nothing about DB's but i want to learn.; after a few attempts i finally got mysql to run (' starting mysql dtabase server') and the next step i am told is to make a pswd. well, i typed 'mysql -u root' to connect to the server like the manual says to do and all i get is 'command not found'.

DECLARE EXIT HANDLER FOR NOT FOUND
It is my understanding the following declaration:

DECLARE EXIT HANDLER FOR NOT FOUND BEGIN SET outStatus = statusNotFound(); END;

in the context of this stored procedure:

CREATE PROCEDURE spfSelectById(OUT outStatusSMALLINT UNSIGNED,
IN inSpfIdSMALLINT UNSIGNED)
BEGIN
DECLARE EXIT HANDLER FOR SQLEXCEPTION SET outStatus = statusSqlException();
DECLARE EXIT HANDLER FOR NOT FOUND BEGIN SET outStatus = statusNotFound(); END;

SELECT spfId, type, description FROM spf WHERE spfId = inSpfId;

SET outStatus = statusSuccess();
END;;

should trigger the execution of the handler if the select does not find a match on spfId; is my understanding correct or flawed?

Innodb_autoinc_lock_mode - Not Found As A System Variable.
I wanted to solve the issue regarding AUTO_INCREMENT resetting itself back to 1 afer a server shutdown. So i did some digging up and read that after mySQL 5.1.22 they started adding a variable to the my.cnf file called

innodb_autoinc_lock_mode

Apparently this variable can store the last value of auto_increment even afer a server shutdown. The problem that im having is that I have mySQL 6.0 (windows). and that variable is not in the my.cnf file. I also checked the manual and this is supposed to be a dynamic variable, meaning i could use SET in the Command line client. I tried this also and it did not work. In fact, when im in the Command Line Client and I enter SHOW VARIABLES;. a list of about 250 variables is displayed but innodb_autoinc_lock_mode is not there.

Im using mySQL 6.0 (for windows) I downloaded the installation program. i did not compile mysql myself.

Big SELECT: Ordering Results By Where Matches Are Found
I'm sure there must be an accepted technique for this, but it's something I haven't tried before, so if anyone can point me in the right direction I'd be grateful.

I'm writing a search facility for a site where the data is stored in several tables - let's say 5 for this example - and I want to order my results according to where (if anywhere) matches are found. So...

Let's say I have tables 'speakers', 'topics', 'speakers_topics', 'articles', 'other'.
'speakers' is a table of speakers, with id, name and some text fields.
'topics' is a list of topics they address
'speakers_topics' relates the above two by pairs of id numbers
'articles' and 'other' are further tables of text data with possibly more than one row for some speakers, identified by id.

I want to search the data in the following order:
name from 'speakers'
topics
text data from 'speakers'
text data from 'articles' and 'other'
...and order the results according to where in that hierarchy a match is found.

So, if the user's search term matches one speaker's name field, another's topic and someone else's text data, that's the order in which the results should be ordered. Also, if the same person is matched from, say, both name and text fields (which is very likely, as their name will almost certainly appear in some of the text), the name should take precedence in the ordering.

To complicate matters further, I'd like if possible to extend this to an and/or situation. If the user enters two or more words, any results that match all the words should be ordered above those that match only some of the words.

I can probably do this relatively easily with a series of separate queries (I'm doing all this from PHP, by the way), but that strikes me as inefficient.

How To Find Duplicates In Two Columns And Count For Each Row Found
I have a little advanced task at hand. I want to search in a table, and find duplicates and count each duplicate found.

Imagine this table

Table | Column Name

Userinfo
Firstname | Lastname

In this table there are many with the same firstname and lastname. I would like an sql statement which find these duplicates and for each duplicate also writes how many occurences it found. So the output I would like is something like this:

John Smith 5
Eva Something 8
and so on.

Deadlock Found When Trying To Get Lock, Try Restarting Transaction.
Im getting the following SQLException during an insert or a delete or an update statement. "Deadlock found when trying to get lock; try restarting transaction message from server: lock wait timeout exceeded; try restarting transaction".

I have a jsp page which does an insert, i have another jsp page which has a delete query that deletes a particular row, and then tries to insert (using an id,which is a primary key), then there is an update statement. In both these pages, i sometimes get the above mentioned exception. When the Tomcat server is restarted, i do not get the exception.

Joining Two Tables And Getting Only Those Results, Where No Matching Entires Have Been Found
I have two tables, "fantasybets" and "fantasybets_placed".

What I want to achieve, is to query "fantasybets" and "fantasybets_placed" to only display those bets, where a given user has paced no bets yet.

Here are examples for my tables:

BETS
______
bet_id
bet_name

BETS_PLACED
______
bet_placed_id
bet_placed_username

SELECT * FROM bets
LEFT JOIN bets_placed
ON bets.bet_id=bets_placed.bet_placed_id
WHERE ((bets_placed.bet_placed_id IS NULL)
OR (bet_placed_username <> "given_username"))

However, this doesn't work as expected, as i get all those bets for a given user, where some other user has placed bets on too. However, I only want those bets, where a defined user "given_username" has NOT placed a bet yet.

MySQLerror: Data Source Name Not Found And No Default Driver Specified
using this connection string;

"DRIVER={MySQL ODBC 3.51 Driver};SERVER=localhost;PORT=3306;DATABASE=dbtest; USER=root;PASSWORD=admin;OPTION=3;"

i still encounter this message:

[Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified

my connection works on the server.. but it doesnt work in clients?

No Mysql Pid File Found. Looked For /var/lib/mysql/localhost.localdomain.pid
The log of my system says:

"no mysql pid file found. looked for /var/lib/mysql/localhost.localdomain.pid"

how to deal with problem?

Returning "binary Flags" With Found Record(s)
I am using the 'binary flag' principle for defining (multiple) modes for an
element..

Elements table:
id name flags
1 a 9
2 b 3
3 c 11

Flags table:
id value name
1 1 mode 1
2 2 mode 2
3 4 mode 3
4 8 mode 4

Now..
Is it possible with one query to get all the modes listed??
to get a result like:
1 a mode1, mode 4
2 b mode 1, mode 2
3 c mode3, mode 4

Query Table A And Return Records Not Found In Table B.
Given table A and table B.  Table A is a master table and table B is a list of authority or reference which is fed by selecting records from table A (via an UI I wrote in PHP).

The interface is working fine but I am left with one problem.    If I need to edit my table of authority, the source table (table A) should return ONLY entries not previously selected.

That said, if the A.key is in B.key, records from table A matching B.key should be omitted.

I could do this by looping through the query results from table A and querying table B, if no match, show A.key.  This I think is a bit antiquated and possibly too involved.

Is there a joint query command I can use?

Log-error File 'not Found' But This Same File Being Written To?!
I recently put up a message board and I get around 150+ hits a day so far. I've recently had some people sign up with my board but unfortunately, no one seems to be posting (they're just lurking around). I'm currently doing some promotions in the hope that others will post (like giving away a Lord of the Rings poster to all users who has 20 posts or more).

Of course, my board has only been up for less than a month (3 weeks to be exact) so I know I need to be patient but I would appreciate any advice you can give me on how to turn lurkers into posters.

PHP Page Displaying When First Hit, Page Not Found On Refresh.
I have installed MYSQL sucessfully and PHP. I usually code in ASP (I know, I know) & SQL Server (Yes OK). And have recently turned to the light!

When I load a page for the first time (On IIS) it looks fine and new data change in the database is displayed and it looks great.

When I click a link to another page with links to the DB then I get Page Cannot be found, if I go back to the URL for the site (thus index.php) Refresh and then click the link it works??!

Last Entry
When i view tables and rows in phpmyadmin, the latest entry (albeit the higher ID number) is on top. Does it work that way inthe internals of mysql? If I select 10 rows with a limit of 10, will i get the latest inserts (assuming the last row id was 49, 49 48 47 46 45... instead of 1,2,3... )?

Duplicate Entry For Key 1
This has me completely stumped, and I've never see anything like it.

Here's my query:

SELECT mus_searchsynonyms.good_term
, mus_searchsynonyms.alt_term
FROM s01_MUS_SearchSynonyms AS mus_searchsynonyms
WHERE mus_searchsynonyms.active = 1
AND mus_searchsynonyms.alt_term = ".quote_smart($search_query)."
AND mus_searchsynonyms.auto_replace = 0
When I use PHP to execute the query, I get the following error message (mysql_error()): (obviously <<$search_query>> is whatever the value of the variable is)

Quote:

Duplicate entry ?-<<$search_query>>' for key 1

When I run the query in phpmyadmin, the query is fine...

First, I don't know why I'd get a key error on a select query, 2nd, there aren't any duplicates in my table...

Most Popular Entry?
I have a quite simple question which I hope can be answered.

Whenever someone downloads a video from my site, a database entry is made with the time and name of the video. I run a php script everyday which discards entries older than 7 days. So I have all that figured out.

What I want to know is, how do I work out what the 10 most popular entries are in the database (by name)?

For example, 8 entries may have the name Video1.wmv, 4 entries may have the name Video4.wmv.etc.

I'm not sure if this can be done via a mysql query, or if php will somehow have to select the most popular entries.

Can't Get Newest Entry
I am trying to select two entries from two databases with one query:

$query = "SELECT ibf_members.id,ibf_members.members_display_name,ibf_member_extra.interests,ibf_member_extra.website,ibf_member_extra.avatar_location,ibf_members.joined FROM ibf_members,ibf_member_extra ORDER BY `joined`,`id` DESC LIMIT 1";
However, it won't show the latest member. It always shows member #1 which is the earliest member. I'm probably doing this wrong. Please help if you can.

Deleting An Entry?
I am using v4.0.20, and I need to figure out how to delete an entry. The database is for a game server I run. It is private, and so I have to add and remove members manually. Here is what I enter to add a person;

INSERT INTO Permissions VALUES ('Sue', 'PRIVATE');

I have never had to remove a player until now! How would I go about removing this entry?

GUI For Data Entry
I am new to MySQL. 10 years ago I was into filemaker development. FileMaker has very nice built-in tools for creating user interface forms for data entry, retreival, editing, etc. So, what tools (preferable for Mac OS X) are available do this with MySQL?

Anyone know of any good apps for taking a MySQL database and automatically creating the PHP code to interact with the database? I've seen one or two for Mac OS X but the Mac OS X solutions seem to be somewhat half-baked compared to the Windows solutions.

Selecting Last Entry
All I need some help i have been trying to write this program for a friend that owns a internet cafe. The DB he uses is in sql he wants people to be able to see their time on his website i almost have it all written except for one problem when it selects there time it brings up all the changes to their account as in if he bought a hour yesterday it would so it i want it to show the current or last entry in the columns heres the code

SELECT userid, username, usertime, datetime
FROM users a, log b
where username = 'Their UserName'
order by datetime;

Is there any command so it shows only the last entry?

Double Entry
Howe can i prevent a double entry in a database.
Users can input their data into table.
But how to prevent a double entry.

Users can enter name and date
So users can enter their names only on one date.

Data Entry Via GUI
Is there a tool for MySQL for creating data-input forms, similar to the FORM VIEW in MS Access, that do not require programming to simply insert data into existing cells or to add rows for new data (and then insert the data)? I looked at the MySQL Browser GUI tool, but didn't see anything about data entry.

Only Allowing One Entry
I am trying to create a simple guetbook. I have it working, but the thing is, it only lets me enter one entry and can't work out why.

This is the insert code i have:

PHP Code:

$insert="Insert into visitordata (name,email, location) values('$name','$email, '$location')";
mysql_query($insert) or die("Could not insert your entry");

Duplicate Entry
I'm currently developing a shopping cart for a local business using php and mysql. I was testing the registration form on the site when all of a sudden I got this error: "Duplicate entry 'tom@thisi.com' for key 1" (that's just a random test email by the way).

I worked out that I only get this error when I use an email that I've used before to signup. Even though I had deleted the previously used email address, it returns an error that it's a duplicate entry. Another strange thing is that the 'email' field is just an ordinary field so a duplicate entry shouldn't cause a problem anyway. If I truncate the table, it's all fixed of course - that's until I use that email address, then delete that account and try and use that email address again.

Match An Entry
Software receives alerts and feeds them into a database table. I need a stored procedure that takes each entry before it enters the table and compares it with the entries already in the table. If the entry about to be added matching one that is already in the table...I need it to cancel each other out and be removed from the database table. If the entry about to be added DOESN'T have a match then it should be added to the table. Is this possible?

Update Entry
Below is the code that I am using. It does the updating for me which is perfect but I cannot get the date part right. It updates all entries. I want it to update any entry that is one day older than presentday.
so update entry if curdat = yesterday.. Please help!

result = mysql_query("UPDATE `shows` SET status=81 WHERE date <= (CURDATE() - INTERVAL 1 DAY)");

Duplicate Entry
I have a table with 10 columns and i insert data into it which is unique in a column called (scrape), so to do that i use
Code:

<?
$reb = select count(*) from table where scrape = '$scrape';
if (mysql_result($reb,0) > 0) {
echo "Item Already Added!<br>";
} else {
add the item
}
?> //I know it's a n00bish way

Validate Entry
How can i validate entries based form Table1 before adding to table2. It means, all entries can be entered into table2 as long as it is already in table1

Table1
id val1
1 002
2 003
3 004

Table2
id val2
1
2
3

Duplicate Entry
I am with a hosting providing use xcart and i keep tgettin this error message:

Quote: INVALID SQL: 1062 : Duplicate entry '' for key 2
SQL QUERY FAILURE: INSERT INTO xcart_products (provider, add_date)
VALUES ('master', '1109438680')

First Letter Of An Entry
Is there a query that allows me to select an item that starts with a specific letter, or a range of letters?
for example, i have:

apple
banana
orange

in my table. Can i issue a query that selects anything that has the first letter in the range a-d?

Duplicate Entry Error ?
I have this table, now i won't it to be possible to have duplicated entrys. Is that possible with this table, or would i have to give it a column with autoincreasment or something?

Mysql, 2nd Highest Entry
how to pick 2nd highest entry from any table in mysql database??
for highest entry i can go by order by desc clause
well limit 0,2 can give me 2 highest entries
but what if i want only 2nd highest entry
any help would be gr8ly appreciated

DELETE DUPLICATE Except LAST ENTRY?
- HOW TO DELETE DUPLICATE FROM MYSQL EXCEPT LAST ENTRY?

CREATE TABLE `chatone` (
`id` int(255) NOT NULL auto_increment,
`username` varchar(30) NOT NULL default '',
`message` varchar(250) NOT NULL default '',
`time` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

--
-- Dumping data for table `chatone`
--

INSERT INTO `chatone` VALUES (1, 'test', 'orange', &#55614;&#57159;-03-13 15:25:24');
INSERT INTO `chatone` VALUES (2, 'test', 'orange', &#55614;&#57159;-03-13 15:25:24');
INSERT INTO `chatone` VALUES (3, 'test', 'apple', &#55614;&#57159;-03-13 15:25:48');
INSERT INTO `chatone` VALUES (4, 'donald', 'hello', &#55614;&#57159;-03-13 15:25:48');

Pulling Out The 2nd Entry From A Table
i have an application already installed. It has an table to handle attachments to articles. The structure is;

attachment_id |question_id| file_path

Is it possible to pull out the 2nd entry in a table for a particular question (using its question_id)?

eg something like?
Select * From attachments WHERE question_id = $question_id Limit ROW 2

Ive search the mysql online manual but im getting really lost!

If Entry Not Found

I want to query my data base for a certain user and id and I want to display a message user not found if the entry isn't there or display the entry if it is.