CODEWITHSUNDEEP
javascriptphpjqueryajax
user3817516
user3817516

Reputation: 49

Receive multiple value from php file via ajax call

Below is my ajax call code. I want to send one data in .php file via ajax call and want to get two values from .php file. This two values I want to set in different 'input' tags whose id are 'course_name' and 'course_credit'.

Here my ajax call return correct value(real value from DB table) of 'course_name' input tag.
But 'MY PROBLEM IS' the value of input tag whose id is 'course_credit' shows 'success'. How can I get the correct value(real value from DB table) of id 'course_credit' ?

I have a 'select' tag which id is 'c_select'

HTML:

<input type="text" name="course_name" id="course_name" value=""/>
<input type="text" name="course_credit" id="course_credit" value=""/>

AJAX :

$('#c_select').change(function(){
    $.ajax({
        type:'post',
        url:'get_course_info_db.php',
        data: 'c_id='+ $(this).val(),                 
        success: function(reply_data1,reply_data2){
            $('#course_name').val(reply_data1);
            $('#course_credit').val(reply_data2);
        }
    }); 
});

get_course_info_db.php

<?php 
       include('db_connection.php'); 
       $c_id = $_POST['c_id'];
       $result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'"); 
       $all_course_data = mysql_fetch_array($result);
       $c_name = $all_course_data['c_name'];
       $c_credit = $all_course_data['c_credit']; 
       echo $c_name,$c_credit;
       exit();  
 ?>

Upvotes: 2

Views: 18937

Answers (4)

t0mpl
t0mpl

Reputation: 5025

success: function(reply_data1,reply_data2){
    $('#course_name').val(reply_data1);
    $('#course_credit').val(reply_data2);
}

second arguement is the status of http request, you have to encode the answer, i suggest you JSON in your php 

$c_credit = $all_course_data['c_credit']; 
echo json_encode(array('name' => $c_name,'credit' => $c_credit));
exit();

and in your javascript

 success: function(response,status){
     var datas = JSON.parse(response);
     $('#course_name').val(datas.name);
     $('#course_credit').val(data.credit);
 }

this is not tested, but this is the way to do it

Upvotes: 2

Domain
Domain

Reputation: 11808

AJAX code:-

$('#c_select').change(function(){
    $.ajax({
        type:'post',
        url:'get_course_info_db.php',
        data: 'c_id='+ $(this).val(),                 
        success: function(value){
            var data = value.split(",");
            $('#course_name').val(data[0]);
            $('#course_credit').val(data[1]);
        }
    }); 
  });

PHP code:-

<?php 
     include('db_connection.php'); 
     $c_id = $_POST['c_id'];
     $result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'"); 
     $all_course_data = mysql_fetch_array($result);
     $c_name = $all_course_data['c_name'];
     $c_credit = $all_course_data['c_credit']; 
     echo $c_name.",".$c_credit;
     exit();   
?>

Upvotes: 8

Raven
Raven

Reputation: 207

The success callback is Function( PlainObject data, String textStatus, jqXHR jqXHR ); http://api.jquery.com/jQuery.ajax/

php: 

$data = array(
    'name' => $c_name,
    'credit' => $c_credit,
);
echo json_encode($data);

javascript:

success: function(data) {
    var result = $.parseJSON(data);
    $('#course_name').val(result.name);
    $('#course_credit').val(result.credit);
}

Upvotes: 3

Hexus
Hexus

Reputation: 350

I'd suggest using JSON to encode the data you fetch from the database.

Try changing your ajax call as follows:

$('#c_select').change(function(){
    $.ajax({
        type:'post',
        url:'get_course_info_db.php',
        data: 'c_id='+ $(this).val(),
        dataType: 'json', // jQuery will expect JSON and decode it for you
        success: function(reply_data){
            $('#course_name').val(reply_data['c_name']);
            $('#course_credit').val(reply_data['c_credit']);
        }
    }); 
});

And your PHP as follows:

include('db_connection.php'); 
// Escape your input to prevent SQL injection!
$c_id = mysql_real_escape_string($_POST['c_id']); 
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'"); 
$all_course_data = mysql_fetch_array($result);
echo json_encode($all_course_data);
exit();

I haven't tested this but I imagine it'd work for you.

Upvotes: 0

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