#23214 [Fbk->NoF]: When exactly is mysql_insert_id() optional?

ID: 23214 Updated by: sniper@php.net Reported By: Stefan_255 at hotmail dot com -Status: Feedback +Status: No Feedback Bug Type: Documentation problem Operating System: win2k PHP Version: 4.3.0 New Comment: No feedback was provided. The bug is being suspended because we assume that you are no longer experiencing the problem. If this is not the case and you are able to provide the information that was requested earlier, please do so and change the status of the bug back to "Open". Thank you. Previous Comments: ------------------------------------------------------------------------ [2003-04-21 16:21:25] alindeman@php.net Status -> Feedback I still think this is a user error, the function works fine... ------------------------------------------------------------------------ [2003-04-20 10:53:18] philip@php.net Please explain what the disclaimers are, we shouldn't rely on user comments as part of the manual. Stefan, some questions: Do you have multiple database connections? Can you create a short short that demonstrates the problem? (one that connects/selects to mysql..). This also might just be some old bug that was fixed as 4.3.0 works. If it was an introduced feature (i.e. resource was made optional) we need to document when that happened. Reopening... ------------------------------------------------------------------------ [2003-04-20 09:58:45] momo@php.net there is enough disclaimers on the site comments. ------------------------------------------------------------------------ [2003-04-14 17:20:55] Stefan_255 at hotmail dot com I copied a few lines of code from the net and got it to work in my own test environment. When I placed the software on the production server (linux PHP 4.1.2) - it only "almost" worked. The problem was that it had lines with statements like: mysql_insert_id(); where strange id's sometimes were returned ! Then, when I changed the code to: mysql_insert_id($con); // The “resource link_identifier” everything was fine. I believe that what happened was that my program received id's from somebody else doing INSERT on the same server? My conclusion was: Never use the function without the "resource link_identifier" ! And then again, - Why should you? In the documentation it says:"If link_identifier isn't specified, the last opened link is assumed." My code was simply: $sql= "INSERT INTO xrefs (referrer) VALUES ('$ref')"; echo "sql $sql <br>"; $result = mysql_query($sql,$con); //$referrer_id = mysql_insert_id(); // FAILED $referrer_id = mysql_insert_id($con); // WORKED OK ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=23214&edit=1