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Warning: mysql_query(): 3 is not a valid MySQL-Link resource

Tags:

php

mysql

I got this odd error and I can't figure out where it came from:

Warning: mysql_query(): 3 is not a valid MySQL-Link resource in (...)

What's up with the 3? I don't get it. Has anyone experienced this error themselves?

like image 813
Pieter Avatar asked May 17 '10 17:05

Pieter


2 Answers

PHP uses resources as a special variable to hold links to external objects, such as files and database connections. Each resource is given an integer id. (Documentation)

Failed Connections

If the database connection fails you'll likely get a "Specified variable is not a valid MySQL-Link resource" error, as Dan Breen mentioned, since the variable that is supposed to hold the resource is null.

$link = mysql_connect('localsoth','baduser','badpass'); // failed connection $result = mysql_query("SELECT 1", $link); // throws error 

Since you're getting a specific resource ID in the error message, the database connection likely closed unexpectedly for some reason. Your program still has a variable with a resource ID, but the external object no longer exists. This may be due to a mysql_close() call somewhere before the call to mysql_query, or an external database error that closed the connection.

$link = mysql_connect(); mysql_close($link); // $link may still contain a resource identifier, but the external object is gone mysql_query("SELECT 1", $link); 

Reusing Connections

An issue with the mysql extension and mysql_connect() is that by default if you pass the same parameters in successive calls, it will re-use the existing connection rather than create a new one (Documentation). This can be fixed by passing true to the $new_link parameter.
I encountered this myself on a test system where the data from two separate databases in production were combined on to one test server, and in testing the mysql_xxx() function calls walked over each other and broke the system.

$link1 = mysql_connect('localhost','user','pass'); // resource id 1 is given $link2 = mysql_connect('localhost','user','pass'); // resource id 1 is given again mysql_close($link2); // the connection at resource id 1 is closed mysql_query("SELECT 1", $link1); // will fail, since the connection was closed 

Using $new_link:

$link1 = mysql_connect('localhost','user','pass'); // resource id 1 is given $link2 = mysql_connect('localhost','user','pass', true); // resource id 2 is given mysql_close($link2); // the connection at resource id 2 is closed mysql_query("SELECT 1", $link1); // the connection at resource id 1 is still open 

Edit:
As an aside, I would recommend using the MySQLi extension or PDO instead, if possible. The MySQL extension is getting pretty old, and can't take advantage of any features past MySQL version 4.1.3. Look at http://www.php.net/manual/en/mysqli.overview.php for some details on the differences between the three interfaces.

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gapple Avatar answered Sep 22 '22 17:09

gapple


I also had this problem. In examining my code I found I had an include of a script that closed the connection, so when php tried to close it again we got the error.

To solve this, just check if the connection is open before trying to close it:

instead of:

mysql_close($con);

Do this:

if( gettype($con) == "resource") {
    mysql_close($con);
}
like image 30
Tarilonte Avatar answered Sep 25 '22 17:09

Tarilonte