Ajax can not give success response

Question

I am fetching subject details of student based on name and admission_year which is getting from ajax call but ajax can not give success response so subject field can not display subject. Plz need solution how to get success message. Is there any wrong in my coding? The whole coding is shown below. script:

<script type="text/javascript">
    $('#branch_name,#class_name').on('change', function(){
    //gender_new = $('#gender_new option:selected').val();

    class_name = $('#class_name option:selected').val();
    branch_name = $('#branch_name option:selected').val();
    name=$('#name').val();
    admission_year=$('#admission_year').val();

    $.ajax({
        type :'POST',
        dataType:'json',
        data : { class_name : class_name, branch_name : branch_name, name : name, admission_year : admission_year },
        url : 'tc_search.php',
        success : function(result){
         $('#subject1').val(result['subject1']);
         $('#subject2').val(result['subject2']);

        }
    });

});
</script>

tc_search.php

<?php
include('connection.php');

$class_name=$_POST['class_name'];
$admission_year=$_POST['admission_year'];
$name=$_POST['name']
$branch_name=$_POST['branch_name'];

$query = "SELECT * FROM admit_senior WHERE name = '$name' and admission_year='$admission_year'";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
echo json_encode($row);

?>

html:

<input type="hidden" class="form-control" id="admission_year" name="admission_year" readonly <?php echo "value='{$row1['admission_year']}' "?>/>

    <div class="form-group">
    <label class="col-sm-3 control-label">Name :</label>
    <div class="col-sm-3">
    <input type="text" class="form-control" id="name" name="name" <?php echo "value='{$row1['name']}' "?>/>
    </div>
    </div>
    <div class="form-group">
    <label class="col-sm-3 control-label">2) He/She appeared/did not appear examination:</label>
    <label class="col-sm-3 control-label">Branch Name:</label>
    <div class="col-sm-3">
    <select id="branch_name" name ="branch_name" class="form-control">
    <option selected disabled readonly>Select branch</option>
    <option value="B.A">B.A</option>
    <option value="B.Com">B.Com</option>
    </select>
    </div>
    </div>

    <div class="form-group">
    <label class="col-sm-3 control-label">Class Name:</label>
    <div class="col-sm-3">
    <select id="class_name" name ="class_name" class="form-control">
    <option selected disabled readonly>Select Class</option>
    <option value="Part-I">Part-I</option>
    <option value="Part-II">Part-II</option>
    <option value="Part-III">Part-III</option>
    </select>
    </div>
    </div>

    <div class="form-group">
    <label class="col-sm-1 control-label">1.</label>
    <div class="col-sm-3">
    <input type="text" class="form-control" name="subject1" id="subject1"/>
    </div>

    <label class="col-sm-1 control-label">2.</label>
    <div class="col-sm-3">
    <input type="text" class="form-control" name="subject2" id="subject2"/>
    </div>

Have you check the network inspector you browser? Does the request return any error? If you are using php 7 you need to wrap $name and $admission_year with {}. — Zenel Rrushi, Sep 27 '19 at 08:04
@leli.1337 basic parsing of variables inside double-quoted strings still works the same in PHP 7, there should be no need to wrap them in { … } — 04FS, Sep 27 '19 at 08:06

devpro · Accepted Answer · 2019-09-27T10:33:55.107

1

First of all mysql_* is deprecated long time ago and closed in PHP 7.

Use mysqli_* or PDO with prepared statement, which will help you to prevent your code with SQL Injection.

Second, use mysqli_num_rows() to check either getting result or not then you can use mysqli_fetch_array.

Issue in your code is that, you are using json_encode() for your result. but fetching result like result['subject2'] which is wrong, you can get the result from json response as like

result.subject2

its better to use console.log(result) and check result in browser console for debugging.

Some useful links:

Are PDO prepared statements sufficient to prevent SQL injection?

How can I prevent SQL injection in PHP?

edited Sep 27 '19 at 10:33

answered Sep 27 '19 at 08:12

devpro

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console.log(result) is not give output send me detail coding – Nikhil korade Sep 27 '19 at 09:24
@Nikhilkorade: what not working.. use this `print_r($row); exit;` before this line `echo json_encode($row);` and share the result – devpro Sep 27 '19 at 09:25
result will be shown in browser console , Ajax's response section @Nikhilkorade – devpro Sep 27 '19 at 09:26
that result is not display in control i think control is not going to tc_search.php – Nikhil korade Sep 27 '19 at 09:32
@Nikhilkorade: check browser console or network tab, where you can see your ajax request – devpro Sep 27 '19 at 09:33
@Nikhilkorade use `alert('here');` after this line in jquery `$('#branch_name,#class_name').on('change', function(){` are u getting alert or not – devpro Sep 27 '19 at 09:38
yes, i get alert with all value that i had send through ajax. – Nikhil korade Sep 27 '19 at 09:40
@Nikhilkorade: if ajax not working then u definitely getting some errors in console. – devpro Sep 27 '19 at 09:43
use these ` ` first option of your dropdowns @Nikhilkorade – devpro Sep 27 '19 at 09:50
@Nikhilkorade: i hope u got the solution – devpro Sep 27 '19 at 10:16
then resolve jquery error first. which you are getting in console @Nikhilkorade – devpro Sep 27 '19 at 11:12
glad to help u @Nikhilkorade – devpro Sep 27 '19 at 12:32
hello need help in ajax success. I want to return multiple result into single result with comm separated of following variable var subject=result.subject2+result.subject3+result.subject4; – Nikhil korade Sep 28 '19 at 07:08
@nikhilkorade no not like that . use each in jquery inside ajax response this will use for loop and I think u also need to use json stringify ... Need to use loop in jquwry – devpro Sep 28 '19 at 08:42

Ajax can not give success response

1 Answers1