when more than one link is opened (to different db) in a php script, if mysql_error() is called without the optional link-id parameter it returns the error status relative to the last opened link whereas documentation states it should be about the last mysql_*() function call.

So far it could be just a documentation problem (not clear enough) yet I'd expect mysql_error() to be smart enough to identify the link used by the most recent mysql_*() function rather just than picking the last created one.

Obviously the same applies for mysql_errno()


doc extract for mysql_error(): 
--------------------
Returns the error text from the last MySQL function, or '' (the empty string) if no error occurred. 
[...]
Note that this function only returns the error text from the most recently executed MySQL function (not including mysql_error() and mysql_errno()), so if you want to use it, make sure you check the value before calling another MySQL function.

Script
----
Here is a simple script to play with to view what I probably made less understandable by writing...
Usage: test.tab1 and guestbook.book db's and tables must be there, the no_*such_table's table should generate an error... altering the order of connections changes the output as well.

<?
 
print "\nDB1\n";
$db1 = mysql_connect("localhost", "user", "user");
print mysql_error();
mysql_select_db("test");
print mysql_error();
 
print "\nDB2\n";
$db2 = mysql_connect("localhost", "guest", "guest");
print mysql_error();
mysql_select_db("guestbook");
print mysql_error();
 
print "\nQUERY1.1\n";
mysql_query("select * from tab1", $db1);
print mysql_error();
 
print "\nQUERY1.2\n";
mysql_query("select * from no_1such_table", $db1);
print mysql_error();
 
print "\nQUERY2.1\n";
mysql_query("select * from book", $db2);
print mysql_error();
 
print "\nQUERY2.2\n";
mysql_query("select * from no_2such_table", $db2);
print mysql_error();
 
print "\nQUERY1.1\n";
mysql_query("select * from tab1", $db1);
print mysql_error();
 
print "\nQUERY1.2\n";
mysql_query("select * from no_3such_table", $db1);
print mysql_error();
 
?>