This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 3 years ago.
Notice: Undefined index: id_desafilter in D:\Sofwares\XAMPP
7.5.5.0\htdocs\project2\views\filterdata.php on line 45
Notice: Undefined index: status_krlfilter in D:\Sofwares\XAMPP
7.5.5.0\htdocs\project2\views\filterdata.php on line 46
that means that your variable may not be defined.
Use isset function.
You should wrap it into a condition like this :
if(!isset($id_desafilter)){
$id_desafilter = "WHAT_YOU_WANT";
}
else{
$id_desafilter = "WHAT_YOU_WANT";
}
This is how you can make sure your variable is define.
But please make sure to join your code to be helped
Have Fun :)
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 3 years ago.
I am looking at my 'WP_DEBUG' errors and the following two errors/notices pop up in a number of PHP files:
Notice: Undefined variable: defaultUI in...
Notice: Undefined variable: compileShortcodeUI in...
I have checked all of the the PHP files and lines specifically and every single one of these refers to this same bit of code:
$compileShortcodeUI .= "<div class='whatInsert whatInsert_".$shortcodeName."'>".$defaultUI."</div>";
What do I need to change to remove these errors?
You have to define the variable initially.
$compileShortcodeUI = ''; // null or any default value
$defaultUI = ''; // null or any default value
$compileShortcodeUI .= "<div class='whatInsert whatInsert_".$shortcodeName."'>".$defaultUI."</div>";
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
I've been getting this error. Any fix??
PHP Notice: Undefined offset: 1 in line 402
if ($argv[1] == '--cover')
{
$Modules = new Modules;
$Modules->cover();
die();
}
The line is if ($argv[1] == '--cover')
An offset is undefined if it doesn't exist in the array.
Try this:
if(isset($argv[1]) && $argv[1] == '--cover'){
$Modules = new Modules;
$Modules->cover();
die();
This error means that the $argv[1] is undefined or null. Make sure this variable is properly assign or have a value before making a condition.
using print_r($argv); before making a condition will help you if the variable have a value or not.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I am getting an undefined index error on this line of code:
switch ($_REQUEST["___p"])
I think I would need to declare the variable. What would I change the above line to?
Either make sure exists beforehand with isset() or just ignore it:
switch (#$_REQUEST["___p"])
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
Notice: Undefined index: n in C:\Users\Marseille\Desktop\Activation
W7\UwAmp\www\quezzer\question.php on line 5
this is a code tiped at question.php
$number = (int) $_GET['n'];
why i have this error! i have not understand!
Since this is a get variable you're using, from what I understand is that you maybe trying to run the script questions.php directly. Therefore no get variables are set, try
localhost/yourfolder/questions.php?n=5
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
I have self written CMS which does not return correct user's IP address with the following:
define("USER_IP", $_SERVER['HTTP_X_FORWARDED_FOR']);
It gives me the following notice:
Notice: Undefined index: HTTP_X_FORWARDED_FOR in /var/www/p00000/data/www/domain.ru/core/maincore.php on line 30
and line 20 = to the line define("USER_IP"...
A few question:
What should I do to get rid of this notice message?
How can I properly get user's IP address?
Thank you for your help!
if (isset($_SERVER['HTTP_X_FORWARDED_FOR'])){
define("USER_IP", $_SERVER['HTTP_X_FORWARDED_FOR']);
} else {
define("USER_IP", $_SERVER['REMOTE_ADDR']);
...