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
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Alive to die - Anant
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2Welcome to Stackoverflow. Please refer stackoverflow.com/help/how-to-ask and stackoverflow.com/help/mcve . Add more details of the issue that you are facing.Share the relevant code snippets you have tried so far. – Kaushal shah Aug 01 '19 at 05:00
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show your code from where and you get id_desafilter – PHP Geek Aug 01 '19 at 05:01
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1no code - no chance of debugging. – Franz Gleichmann Aug 01 '19 at 05:08
1 Answers
0
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 :)

Lucas
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