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
            Asked
            
        
        
            Active
            
        
            Viewed 45 times
        
    -6
            
            
         
    
    
        Alive to die - Anant
        
- 70,531
- 10
- 51
- 98
- 
                    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
- 
                    show your code from where and you get id_desafilter – PHP Geek Aug 01 '19 at 05:01
- 
                    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
        
- 73
- 1
- 8
