PHP - get variable obtained by ajax

Question

I used ajax to send to me a php variables in this way:

<script type="text/javascript">
$(document).ready(function(){
    $("#disdiciButton").click(function(){

       var log_user = <?php echo json_encode($log_user);?>;
            alert(log_user);

            $.ajax({
               type: 'POST',
               url: 'disdici.php',
               data:log_user,
               success: function(data) {
                   $("#deleteResponse").text(data); 

               }
        });
     });
});
</script>

Now I'm trying to collect this data in the php page disdici.php but I don't know how. I've tried in this way:

$log_user = "";
if(isset($_POST['data'])){
   $log_user = json_decode(data);         
}
echo 'user: '.$log_user;

but $log_user remain empty. How can I do?

data isn't a posted field. It's a constant in this case, so $log_user will be empty. — Tyr
– Tyr, Commented Jun 23, 2018 at 16:59

Ikki_Phenix · Accepted Answer · 2018-06-23 17:27:29Z

1

You have to write your data like :

data : 'var1=' + value1 + '&var2=' + value2 + '&var3=' + value3...

So, you can get the variable in php :

$_POST['var1'] = value1; 
...

In your example :

<script type="text/javascript">
$(document).ready(function(){
   $("#disdiciButton").click(function(){

      var log_user = <?php echo json_encode($log_user);?>;
           alert(log_user);

           $.ajax({
              type: 'POST',
              url: 'disdici.php',
              data:'log_user=' + log_user,
              success: function(data) {
                  $("#deleteResponse").text(data); 

              }
       });
    });
});
</script>

In PHP file, you get the value :

$_POST['log_user']

edited Jun 23, 2018 at 17:27

answered Jun 23, 2018 at 17:18

Ikki_Phenix

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