使用each（）打印数组时出现jquery ajax（）问题

Question

In my code I am returning an php array containing records of 7 Students. With jquery ajax() I want to print these records on success function.

DB table Students

+---------------------------+
| name | fathername | Email |
+---------------------------+

submit.php

$query=mysql_query("SELECT * from Students LIMIT 0,6");

   $row= array();
   $row=mysql_fetch_array($query);
   return json_encode($row);

index.php

<script>
    $(function(){
        $("#form1").submit(function(event){
            event.preventDefault();

            $.ajax({
                    url:'submit.php',
                    type:'GET',
                    data:$(this).serialize(),
                    success:function(result){
                $.each(result,function(){
                $('.StudentName').text(result["name"]);
                $('.FatherName').text(result["fathername"]);
                $('.Email').text(result["email"]);
                    });

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

<div class="StudentName"></div>
<div class="FatherName"></div>
<div class="Email"></div>

EDIT I tried to return only 1 result from php and it works i.e.

echo json_encode(mysql_fetch_array($query));

When I return all 6 records the jquery function dont execute i.e.

while($result=mysql_fetch_array($query))
{
echo json_encode($result);
}

Answer 1

There's difference between PHP arrays and JS arrays, you can't simply pass the PHP array to your javascript, so instead you should first json_encode it and send it to js.

This will convert your PHP array to JSON array, eg:

 array(3) {
  [0]=>
    string(3) "foo"
  [2]=>
   string(3) "baz"
  [3]=>
   string(5) "blong"
}

to

string(33) "{"0":"foo","2":"baz","3":"blong"}"

So try -

return json_encode($row);

and then when you catch the response, use parseJSON:

result = jQuery.parseJSON(result);
$.each(result,function(){
   $('.StudentName').text(result.name);
   $('.FatherName').text(result.fathername);
   $('.Email').text(result.email);
});

Edit:

Another thing, instead of return json_encode($row); write echo json_encode($row);

Edit 2

(to send all 6 records)

$final = array();
while($result=mysql_fetch_array($query))
{
    array_push($final,$result);
}
echo $final;