Update multiple checkbox value in php

This question has probably been asked multiple times but I still cannot figure out a proper working solution to my problem. So, I have an sql table (id-> int and status-> tiny-int) which needs to be activated without submit button and retain their current state either on or off. I found the solution to my problem here. But, this is only for one checkbox and I need it to be for four checkboxes. My attempt at it sorta works but I feel there is probably a better way of doing it. Can someone please provide me with any suggestions?

thanks

  


  >
 


  
if (isset($_POST['checkbox1'])) {
         $sql1="UPDATE switch SET status = '1' WHERE id = '1'";
        
     } else {
         $sql1="UPDATE switch SET status = '0' WHERE id = '1'";
        
     }
     
       $result=$conn->query($sql1);
       
     if (isset($_POST['checkbox2'])) {
         $sql2="UPDATE switch SET status = '1' WHERE id = '2'";
        
     } else {
         $sql2="UPDATE switch SET status = '0' WHERE id = '2'";
        
     }
     
       $result=$conn->query($sql2);

asked Dec 26, 2021 at 11:36

1

You only need to multiply your code blocks following @Kudzė 's instructions. Like this:

+Tip: at the future when you want to store inputs from the users in the database please use prepared SQL statements this is prevent SQL injection.

 0){
    if (isset($_POST['checkbox1'])) {
        $sql1="UPDATE switch SET status = 1 WHERE id = 1";
    } else {
        $sql1="UPDATE switch SET status = 0 WHERE id = 1";
    }
    $result=$conn->query($sql1);

    if (isset($_POST['checkbox2'])) {
        $sql2="UPDATE switch SET status = 1 WHERE id = 2";
    } else {
        $sql2="UPDATE switch SET status = 0 WHERE id = 2";
    }
    $result=$conn->query($sql2);

    if (isset($_POST['checkbox3'])) {
        $sql3="UPDATE switch SET status = 1 WHERE id = 3";
    } else {
        $sql3="UPDATE switch SET status = 0 WHERE id = 3";
    }
    $result=$conn->query($sql3);

    if (isset($_POST['checkbox4'])) {
        $sql4="UPDATE switch SET status = 1 WHERE id = 4";
    } else {
        $sql4="UPDATE switch SET status = 0 WHERE id = 4";
    }
    $result=$conn->query($sql4);
}

?>



    
    
    
    


    


        
            Checkbox 1
            >
        

        
            Checkbox 2
            >
        

        
            Checkbox 3
            >
        

        
            Checkbox 4
            >
        

    

answered Dec 26, 2021 at 12:50

adampwebadampweb

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1

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