CODEWITHSUNDEEP
phpjquerysqlajax
Stoppie Chandru DiGi
Stoppie Chandru DiGi

Reputation: 51

Jquery append option in select tag

Here i tried as much i can but couldn't get results. please any guys help me out. i cannot get the correct result using these codes.may be something i'm missing.

$get_medical_aod_name = $_POST['name'];
$select_query = mysql_query("SELECT id, name FROM insurance_companies WHERE cat=$get_medical_aod_name");
while ($row = mysql_fetch_array($select_query)) {
  $result[] = array(
    'id' => $row['id'],
    'name' => $row['name']
  );
}
header('Content-Type: application/json', true);
echo json_encode($result);

and in my script.js 

$('#medical_aid_name').change(function(){
       //alert("working");
       var name = $(this).val();
       $.post('ins_bene.php',
    {'name' : name },
    function(data){
        var select = $('#benefit').empty();
        $.each(data.values, function(i,data) {
            select.append( '<option value="'
                                 + data.id
                                 + '">'
                                 + data.name
                                 + '</option>' ); 
        });
    }, "json");
   });

and i feel everything fine but i'm getting no output :(

Upvotes: 1

Views: 2538

Answers (5)

Nikhil Mohan
Nikhil Mohan

Reputation: 891

Change this in script.js

$.each(data.values, function(key, value)

to

$.each(data, function(key, value) {

OR

Assign a key values to $result

while ($row = mysql_fetch_array($select_query)) {
  $result['values'] = array(
    'id' => $row['id'],
    'name' => $row['name']
  );
}

Upvotes: 0

kongaraju
kongaraju

Reputation: 9596

this should work

while ($row = mysql_fetch_array($select_query)) {
  $result['values'] = array(
    'id' => $row['id'],
    'name' => $row['name']
  );
}

Upvotes: 0

user2334807
user2334807

Reputation:

Instead of the following code:

 var select = $('#benefit').empty();
        $.each(data.values, function(i,data) {
            select.append( '<option value="'
                                 + data.id
                                 + '">'
                                 + data.name
                                 + '</option>' ); 
        });

Use the below code:

    var select = $('#benefit').empty();
    $.each(data.values, function(i,data) {
        $('#benefit').append( '<option value="'
                             + data[i].id
                             + '">'
                             + data[i].name
                             + '</option>' ); 
    });

Here select var don't have the object of the select tag that's why.

Upvotes: 0

Lakshmana Kumar
Lakshmana Kumar

Reputation: 1239

Try this... UPDATED

    $('#medical_aid_name').change(function() {
        var name = $(this).val();
        $.post('ins_bene.php', { 'name': name },
        function(data) {
            alert(data)  // [object][object],[object][object]
            $('#benefit').empty();
            $.each(data, function(key, value) {
                $('#benefit').append('<option value="' + value.id + '">' + value.name + '</option>');
            });
        }, "json");
    });

Upvotes: 1

Farkie
Farkie

Reputation: 3337

You need to quote the query:

$select_query = mysql_query("SELECT id, name FROM insurance_companies WHERE cat=$get_medical_aod_name");

to

$select_query = mysql_query('SELECT id, name FROM insurance_companies WHERE cat="' . mysql_real_escape_string($get_medical_aod_name) .'"');

Please don't use mysql_ functions. Move to PDO!

Upvotes: 1

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