How to copy json array variable data to php variable?

Question

This is a web page for student results and 3 fields data i.e stu_class, section, exam is sending to controller and a json array data variable is coming back from controller to this page now i wanted to show that json variable's data on the webpage. So help me how to do so?

$(document).ready(function(){

$('#examination').change(function(){

  var exam = $('#examination').val();
  $('#exam').val(exam);
  });

$('#enter').click(function(){
 var stu_class = $('#stu_class').val();
 var section = $('#section').val();
 var exam = $('#examination').val();

  $.ajax(
     {
       type : "POST",
       dataType: "json",
       url : "<?php echo base_url('index.php/marksheet/student_result_database'); ?>",

       data : {
              stu_class : stu_class,
              section : section,
              exam : exam
          },
       success: function(stu_data)
        {
          var items = [];
          $.each( stu_data, function( key, val ) {   
               items.push( "<li id='" + key + "'>" + val['student'] + "</li>" );
             });

             alert(items);

          }
       });
      });

});

And i need to copy the data of stu_data variable of jquery to php variable $stu_data so that i can run a loop to show the values in below fields like this-

<?php foreach($stu_data as $data): 
echo $data['id']; 
echo $data['student'];

Something like this. So programmers please help me..!

stu_data is data you just got from PHP. It isn't generated by the client. Why not use it before you send it back? — Quentin
– Quentin, Commented Aug 27, 2014 at 10:42
"to show the values" — You're using Ajax. The data is coming back to your JS code, not to the HTML page. Why aren't you just adding them to the DOM with JS? You're alreadying generating HTML from them. — Quentin
– Quentin, Commented Aug 27, 2014 at 10:43
You might want to check out: stackoverflow.com/questions/19886553/… — designcise
– designcise, Commented Aug 27, 2014 at 10:46

lshas · Accepted Answer · 2014-08-27 10:48:05Z

1

json_decode($str, true) -> turns a json-object into a php assoc array.

json_decode($str, false) -> turns a json-object into a php object.

answered Aug 27, 2014 at 10:48

lshas

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1 Comment

DarkBee Over a year ago

The false is not necessary as the optionals parameters default state is false

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