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Bug #2843 Cannot use variable after a mysql_fetch_array
Submitted: 1999-11-27 13:05 UTC Modified: 2000-03-18 09:33 UTC
From: yzhang at sfu dot ca Assigned:
Status: Closed Package: Misbehaving function
PHP Version: 4.0 Beta 3 OS: Linux
Private report: No CVE-ID: None
 [1999-11-27 13:05 UTC] yzhang at sfu dot ca
Assuming $qid is a valid result of a mysql_query:

while ($r = mysql_fetch_array($qid)) {
    ...do something...
}

$r["foo"] = "bar";
echo $r["foo"];

Will not display anything.  $r["foo"] is never assigned the value "bar".  Instead, I must forcefully unset($r); or $r = array; before trying to make the $r["foo"] = assignment.

This works fine in PHP 3 but not in 4.0 beta 3.

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 [2000-03-18 09:33 UTC] zeev at cvs dot php dot net
Fixed.  Thanks for the report!
 [2023-10-10 09:09 UTC] git@php.net
Automatic comment on behalf of derickr
Revision: https://github.com/php/doc-en/commit/a11c5b53bc132289a0ddcc6d3efee5d97f0027b4
Log: Fixed #2843: Error section mistakingly called the previous situation 'Exception' instead of warning
 
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