ID: 23214
Updated by: sniper@xxxxxxx
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:
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[2003-04-21 16:21:25] alindeman@xxxxxxx
Status -> Feedback
I still think this is a user error, the function works fine...
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[2003-04-20 10:53:18] philip@xxxxxxx
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@xxxxxxx
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
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/?id=23214&edit=1
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