Ajax call not getting a success response

Question

We have 2 text fields. We have tried to post user_Name and Password through an ajax call.

We have used PHP services and the username and password do save to the DB successfully.

But we do not get any response from our success or fail.

The code:

PHP:

 if(isset($_GET['ecovauserName']) && isset($_GET['ecovauserage']) ){
    $ecovauserName             = $_GET['ecovauserName'];
    $ecovauserage              = $_GET['ecovauserage'];



     $sql  = "SELECT ecovauserName,ecovauserage FROM ecovauserinfo WHERE ecovauserName = '" . $ecovauserName . "' and ecovauserage = '" . $ecovauserage . "'";
    $query      = mysql_query($sql);

    if (mysql_num_rows($query) > 0)
    {
        echo "fail to post";
    }
    else
        {  echo("Entered into DB inserting the values");

        $insert = "INSERT INTO ecovauserinfo (ecovauserName,ecovauserage)
                        VALUES ('" . $ecovauserName . "','" . $ecovauserage . "')";

        $query  = mysql_query($insert);
        echo $insert;
        if ($query)
        {
            echo "EventFile Successfully stored";
        }
            else
            {


                echo "Insert failed";
            }
        }
     }

Ajax Call:-

 $.ajax({
               url:'http://192.168.3.134:8080/ekova/postevent.php',
               type:'POST',
               data:{ecovauserName :username,ecovauserage:password},
               contentType: "application/json; charset=utf-8",
                dataType: "jsonp",
               success:function(responsive){
                alert(responsive);
               },
               error:function(w,t,f){
                 console.log(w+' '+t+' '+f);
               }
            });

The above code is working fine. The username and password are successfully stored in the DB. But we need to get a success of fail response.

My success:function is not called and so my alert box never runs to notify me of the success.

Please guide me with what is wrong in the code.

**Danger**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Mar 06 '15 at 10:40

score 1 · Answer 1 · edited Dec 16 '16 at 13:57

1

The data type should read "json" not "jsonp"

I would suggest that you use "text" instead of "json" since you are returning normal text results and not json encoded data.

edited Dec 16 '16 at 13:57

Philip Kirkbride

21,381
38
125
225

answered Mar 06 '15 at 10:22

Schalk Keun

538
4
13

thanks for reply.We tried as you said but when we placed JSON ,data Not saved in DB – Pavan Alapati Mar 06 '15 at 10:27
the dataType is used to tell ajax what data is expected to RETURN and not SEND. So this change should not have affected anything in your posting/saving. Also NOTE> you are telling AJAX to POST but you are checking the $_GET globals. You will need to use $_POST['ecovauserName'] and $_POST['ecovauserage'] – Schalk Keun Mar 06 '15 at 10:35

score 1 · Answer 2 · edited May 23 '17 at 12:28

1

You could try removing the line dataType: "jsonp" - that's had some success.

See this post: Ajax success event not working

EDIT - try putting basic console.log statements in both success and fail just to see if either are being hit.

$.ajax({
           url:'http://192.168.3.134:8080/ekova/postevent.php',
           type:'POST',
           data:{ecovauserName :username,ecovauserage:password},
           contentType: "application/json; charset=utf-8",
            dataType: "jsonp",
           success:function(){
            console.log('in success');
           },
           error:function(){
             console.log('in error');
           }
        });

Check whether you're getting an error:

error: function(xhr, status, error) {
  var error1 = eval("(" + xhr.responseText + ")");
  console.log(error1.Message);
  console.log(geturl.getAllResponseHeaders());
  alert("error!"+ geturl.getAllResponseHeaders());
}

edited May 23 '17 at 12:28

Community

1
1

answered Mar 06 '15 at 10:24

hlh3406

1,382
5
29
46

we tried remove jsonp.but success method not calling – Pavan Alapati Mar 06 '15 at 10:31
If you put a console.log("blablabla") into success and fail are either of those called? - answer edited to reflect. – hlh3406 Mar 06 '15 at 10:35
in success function console.log not called but values stored in db please guide me how call success – Pavan Alapati Mar 06 '15 at 10:40
Can you try changing the dataType to "text"? And also put into the success as well as the console.log the following: return alert("Testing"); – hlh3406 Mar 06 '15 at 10:44
we placed case1) dataType:Text we get 'in fail',case2)dataType:json we get 'in fail' case3) dataType:jsonp we get 'in fail' but only case3 data save to DB – Pavan Alapati Mar 06 '15 at 10:51
Could be your headers - you need to print out the error your getting to the console log. Answer edited. – hlh3406 Mar 06 '15 at 10:55
we print console.log(error1.Message); we got this Uncaught TypeError: Cannot read property 'Message' of undefined and alert is not work – Pavan Alapati Mar 06 '15 at 11:02
Ok definitely think its your headers - put this at the top of your PHP file underneath the first – hlh3406 Mar 06 '15 at 11:12
we got error this [object Object] parsererror Error: jQuery111002164177882950753_1425642125857 was not called – Pavan Alapati Mar 06 '15 at 11:43
Moved these messages to chat. – hlh3406 Mar 06 '15 at 11:58

score 1 · Answer 3 · answered Mar 06 '15 at 11:03

Why don't you use status codes returned by the call, as an example:

$.ajax({
  url:'http://192.168.3.134:8080/ekova/postevent.php',
  type:'POST',
  data:{ecovauserName :username,ecovauserage:password},
  contentType: "application/json; charset=utf-8",
  dataType: "jsonp",
  statusCode: {
    404: function () {
      //error
    },
    200: function (data) {
      //response data
    }
  }
});

score 0 · Answer 4 · answered Mar 06 '15 at 11:01

0

Return a json_encode response in your php call, like this:

echo json_encode(array('status' => 'in success'));

answered Mar 06 '15 at 11:01

Juan de Parras

768
4
18

Bang · Answer 5 · 2015-03-06T12:51:23.953

You seems doing a query with cross-domain and jsonp make that possible.

So, in your php code you have to return a jsonp format :

$response = array('success' => false, 'message' => '');

if (isset($_GET['ecovauserName']) && isset($_GET['ecovauserage'])) {
    $ecovauserName = $_GET['ecovauserName'];
    $ecovauserage = $_GET['ecovauserage'];

    $sql = "SELECT ecovauserName,ecovauserage FROM ecovauserinfo WHERE ecovauserName = '" . $ecovauserName . "' and ecovauserage = '" . $ecovauserage . "'";
    $query = mysql_query($sql);

    if (mysql_num_rows($query) > 0) {
        $response['message'] = 'fail to post';
    } else {
        $insert = "INSERT INTO ecovauserinfo (ecovauserName,ecovauserage) VALUES ('" . $ecovauserName . "','" . $ecovauserage . "')";

        $query = mysql_query($insert);
        if ($query) {
            $response['message'] = 'EventFile Successfully stored';
            $response['success'] = true;
        } else {
            $response['message'] = 'Insert failed';
        }
    }
}
// The header is for telling that you are sending a json response
header('content-type: application/json; charset=utf-8');

// formatting a response as jsonp format.
echo $_GET['callback'] . '(' . json_encode($response) . ')';

In your javascript :

$.ajax({
    url: 'http://192.168.3.134:8080/ekova/postevent.php',
    type: 'POST',
    data: {ecovauserName: username, ecovauserage: password},
    contentType: "application/json; charset=utf-8",
    dataType: "jsonp",
    success: function (response) {
        alert(response.message);
    },
    error: function() {
        console.log('error');
    }
});

More information at http://www.geekality.net/2010/06/27/php-how-to-easily-provide-json-and-jsonp

thanks for reply we are new to PHP please explain where we placed two lines header('content-type: application/json; charset=utf-8'); echo $_GET['callback'] . '(' . json_encode($response) . ')'; — Pavan Alapati, Mar 06 '15 at 11:10
add for debug `error: function (jqXHR, textStatus, errorThrown) { console.log(jqXHR); console.log(textStatus); console.log(errorThrown); }` — Bang, Mar 06 '15 at 13:10
we got errors this Object parsererror Error: jQuery111002985340391751379_1425647499106 was not called file:///D:/SaveApp/SaveApp/assets/ForLL/jquery.mobile-1.4.2.min.map Failed to load resource: net::ERR_FILE_NOT_FOUND — Pavan Alapati, Mar 06 '15 at 13:15

Ajax call not getting a success response

5 Answers5