ID: 10887
User Update by: [EMAIL PROTECTED]
Old-Status: Feedback
Status: Open
Bug Type: Unknown/Other Function
Operating system: Windows 2000 Pro
PHP Version: 4.0.5
Description: Variables from forms being undefined
This function did not work in PHP4.05:
elseif ($delete) {
// delete a record
$sql = "DELETE FROM models WHERE model_id=$model_id";
$result = mysql_query($sql);
echo "$sql Record deleted!<p>";
}
Unless I added isset() like so:
elseif (isset($delete)) {}
But as I added that in other functions that returned a value from mySQL, every value
returned would be "1" or TRUE.
Previous Comments:
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[2001-05-15 16:23:28] [EMAIL PROTECTED]
This script shows no isset(), please be a little more descriptive.
Derick
---------------------------------------------------------------------------
[2001-05-15 16:01:31] [EMAIL PROTECTED]
Here is an example of a script that returned the error in 4.05, and works in 4.04.
elseif ($delete) {
// delete a record
$sql = "DELETE FROM models WHERE model_id=$model_id";
$result = mysql_query($sql);
echo "$sql Record deleted!<p>";
---------------------------------------------------------------------------
[2001-05-15 15:54:10] [EMAIL PROTECTED]
Please post a short reproducing script. (short is under 10 lines)
---------------------------------------------------------------------------
[2001-05-15 15:50:01] [EMAIL PROTECTED]
When using PHP4.05 my script always told me a variable from a form I had on the page
was undefined. Using isset() the script ran without problems, but returned every query
with "1" (my best guess is that is TRUE from isset()). I installed PHO4.04 and
everything ran perfectly.
Mike Peterson
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Full Bug description available at: http://bugs.php.net/?id=10887
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