php return sends the content of included file to javascript

Question

I'm facing a problem in getting the return value from PHP to JSON.

If success or failure the alert will be displayed with the content of a javascript file inside include '../core/init.php';. But I need only success or failed in alert. Please help

Requestfile:

$id = intval($_REQUEST['id']);
$empid=$_REQUEST['empid'];

include '../core/init.php';

if($user_data['empid']==$empid){
    $delete_sql = "delete from datatable where sl=$id";
    mysql_query($delete_sql);  
    echo json_encode("success");
    }
    else{
  echo json_encode("failure");
}

Main Page:

$.ajax({
url: 'returnpage.php',
type: 'post',
data: {
'id' : row.sl,
'empid':row.empid                   
},
success:function(status){
alert(status);
},
error:function(status){
alert(status);
}

Alert window:

<style>

/*html {

    background: url(includes/background.jpg) no-repeat center center fixed; 

  -webkit-background-size: cover;

  -moz-background-size: cover;

  -o-background-size: cover;

  background-size: cover;

}*/

</style>



<script type="text/javascript" src="css/js/browsercheck.js"></script>

 <script >

var browser_name=browserinfo();

//alert(browser_name);

if(browser_name!=="Firefox"){
   window.location="non_compatible_browser.php";
}
</script>



<script type="text/javascript" src="css/js/browsercheck.js"></script>

<script >

   // alert("Application Will be down..!!\nFrom 23-Jan-2016 07:15 PM to 23-Jan-2016 08:15 PM \nPlease save the data. ")

    var browser_name=browserinfo();

    //alert(browser_name);

    if(browser_name!=="Firefox"){

        window.location="non_compatible_browser.php";

    }

    //window.location="Maintenance_page.php";

//exit();

</script>

"failure"

[Don't use `mysql_query`](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — Michał Perłakowski, Jan 13 '16 at 10:55
json_encode should return the array which can be converted as javascript object try like this `json_encode(array('status'=>'success'))` — Sundar, Jan 13 '16 at 10:57
what is the "failure" string int the alert window page then ? — Sn0opr, Jan 13 '16 at 10:57
if you returning json you can mention the return type of ajax as `dataType:'json'` — Sundar, Jan 13 '16 at 10:57
You are most likely generating some output somewhere else, which is delivered together with your result. — Burki, Jan 13 '16 at 10:58
even I tried the same. However it return the above mentioned dump of js and a array at the end. — Mr.Kiran, Jan 13 '16 at 11:01
If I understand, you have JS inside init.php .. Why? It should be in view. But anyway if you use output buffer you can call ob_clean() before echo json_encode("success"); — David Kuna, Jan 13 '16 at 11:03

score 0 · Answer 1 · answered Jan 13 '16 at 10:58

0

use exit after echo as below :

    $id = intval($_REQUEST['id']);
    $empid=$_REQUEST['empid'];

    include '../core/init.php';

    if($user_data['empid']==$empid){
        $delete_sql = "delete from datatable where sl=$id";
        mysql_query($delete_sql);  
        echo json_encode("success");
exit;
        }
        else{
      echo json_encode("failure");
exit;
    }

answered Jan 13 '16 at 10:58

Drudge Rajen

7,584
4
23
43

While this may or may not help (depending on where the unwanted output comes from), an explanation why it should might be a good idea here. – Burki Jan 13 '16 at 11:00
@Burki if anything is echoed in the ajax function call then it will give a same whole page as response . – Drudge Rajen Jan 13 '16 at 11:03

score 0 · Answer 2 · answered Jan 13 '16 at 11:03

Add dataType: 'JSON' to your AJAX JS code:

$.ajax({
url: 'returnpage.php',
type: 'post',
dataType: 'JSON',
data: {
'id' : row.sl,
'empid':row.empid                   
},
success:function(status){
alert(status);
},
error:function(status){
alert(status);
}

Sundar · Answer 3 · 2016-01-13T12:01:44.200

0

JavaScript Changes make sure your post parameters have the values

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<script>
$(document).ready(function(){

    $.ajax({
        url: 'returnpage.php',
        type: 'post',
        data: {
            'id' : row.sl,
            'empid':row.empid                   
        },
        dataType:'json',
        success:function(result){
            alert(result.status);
        },
        error:function(result){
            alert(result);
        } 
    });
});
</script>

PHP changes

<?php

$id = intval($_REQUEST['id']);
$empid=$_REQUEST['empid'];

$data = array();
$data['status'] = 'failure';

//include '../core/init.php';

if($user_data['empid']==$empid){
    $delete_sql = "delete from datatable where sl=$id";
    $result = mysql_query($delete_sql);

    if($result) {
        $data['status'] = 'success';
    }
}
echo json_encode($data);

edited Jan 13 '16 at 12:01

answered Jan 13 '16 at 11:04

Sundar

4,580
6
35
61

Sundar. I'm getting '[object Object]' but not value – Mr.Kiran Jan 13 '16 at 11:33
try this `alert(status.status)` and enable the commented line too – Sundar Jan 13 '16 at 11:40
try now it's got changed monitor the request and response in firebug/network console variable named as result in javascript – Sundar Jan 13 '16 at 12:02
what you getting.. have you checked in your network/console. for me it's working fine no issues – Sundar Jan 13 '16 at 12:43

php return sends the content of included file to javascript

3 Answers3