CODEWITHSUNDEEP
javascriptphpjquerymysqlajax
vinod jaiswal
vinod jaiswal

Reputation: 89

update query when check the checkbox with ajax

I have a checkbox. if I check or uncheck the check box then a update query should be run with Ajax call.

If checks the checkbox then 1 should be update and if uncheck then 0 should be update. But it is not working.

Checkbox is:

<input type="checkbox" id="test_done" value="<?php echo $row1['id']; ?>" <?php if(($row1['test_done'])=='1') echo "checked='checked'"; ?> data-toggle="checkbox">

In above $row1['id'] and $row1['test_done'] are columns name, which are coming from another select query.

Ajax Is:

<script type="text/javascript">
  $(document).ready(function(){
    $("#test_done").click(function(){    
    var val = $(this).val();
         if($(this).is(':checked')){
         $.ajax({ type: "POST", 
         url: "test_done.php", 
         data: {val:val,apply:'1'} 

         });
         }
         else
         {
         $.ajax({ type: "POST", 
         url: "test_done.php", 
         data: {val:val,apply:'0'} 

         });

         }
      });
   });
 </script>

test_done.php is:

include_once("connection.php");
if($_POST['apply']=='1')
{
$val = $_POST['val'];   
 mysql_query("update patient_package_details set test_done='1' where id='$val'") or die(mysql_error());
}
else
{
$val = $_POST['val'];   
mysql_query("update patient_package_details set test_done='0' where id='$val'") or die(mysql_error());
}

Upvotes: 2

Views: 4642

Answers (1)

Pankaj Makwana
Pankaj Makwana

Reputation: 3050

You can use change event instead of click event on checkbox. Check if checkbox is checked then send 1 otherwise 0. on php side as we are sending both values, so you will get value using post. You don't need to check value for $_POST["apply"]. I have set the value of apply variable, so only one query will work.

$("input:checkbox").on("change", function () {
    var val = $(this).val();
    var apply = $(this).is(':checked') ? 1 : 0;
    $.ajax({type: "POST",
        url: "test_done.php",
        data: {val: val, apply: apply}
    });
});

test_done.php:

include_once("connection.php");
$val = $_POST['val'];   
$apply = $_POST['apply'];   
mysql_query("update patient_package_details set test_done='$apply' where id='$val'") or die(mysql_error());

Upvotes: 4

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