jquery ajax function not working

Question

my html goes like this ,

<form name="postcontent" id="postcontent">
    <input name="postsubmit" type="submit" id="postsubmit" value="POST"/>
    <textarea id="postdata" name="postdata" placeholder="What's Up ?"></textarea>
</form>

The jquery code is as follows

$("#postcontent").submit(function(e) {
    $.ajax({
        type:"POST",
        url:"add_new_post.php",
        data:$("#postcontent").serialize(),
        beforeSend:function(){
            $(".post_submitting").show().html("<center><img src='images/loading.gif'/></center>");
        },success:function(response){   
            //alert(response);
            $("#return_update_msg").html(response); 
            $(".post_submitting").fadeOut(1000);                
        }
    });
});

When I click on the submit button , my ajax request is not working , it looks as if the control is being passed to the JQuery submit function , but the ajax request is not executing/working properly,what is wrong ?

There is no error , it's just that the ajax call is not working — Kiran Gopalakrishnan, Feb 15 '14 at 11:44
@KiranGopalakrishnan how you solved it ? is still not working ? — alicef, Jan 29 '16 at 08:59
@KiranGopalakrishnan how you solved it ? would be nice if you share the solution... — alicef, Feb 03 '16 at 00:56

score 13 · Accepted Answer · answered Feb 15 '14 at 11:51

put the event handler function inside $(document).ready(function(){...}). it shall work now

also add preventDefault() to restrict page refreshing

$(document).ready(function() {

            $("#postcontent").submit(function(e) {
                e.preventDefault();
                $.ajax({
                    type : "POST",
                    url : "add_new_post.php",
                    data : $("#postcontent").serialize(),
                    beforeSend : function() {
                          $(".post_submitting").show().html("<center><img src='images/loading.gif'/></center>");
                    },
                    success : function(response) {
                        alert(response);
                        $("#return_update_msg").html(response);
                        $(".post_submitting").fadeOut(1000);
                    }
                });
                e.preventDefault();
            });

        });

I do type the handler inside $(document).ready(); but still not working. Any idea? — Hendry Tanaka, Oct 14 '14 at 05:18

score 4 · Answer 2 · answered Feb 15 '14 at 11:41

try this code

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<Script>
$(document).ready(function(){

$("#postcontent").click(function(e) {

        $.ajax({type:"POST",url:"add_new_post.php",data:$("#postcontent").serialize(),beforeSend:function(){
            $(".post_submitting").show().html("<center><img src='images/loading.gif'/></center>");
        },success:function(response){   
                    //alert(response);
                    $("#return_update_msg").html(response); 
                    $(".post_submitting").fadeOut(1000);                
            }
        });
   });
});
   </script>

<form name="postcontent" id="postcontent">
              <input name="postsubmit" type="button" id="postsubmit" value="POST"/>
              <textarea id="postdata" name="postdata" placeholder="What's Up ?"></textarea>
 </form>

score 2 · Answer 3 · answered Feb 15 '14 at 11:41

2

you need to prevent the default behavior of your form when submitting

by adding this:

$("#postcontent").on('submit' , function(e) {

  e.preventDefault();

  //then the rest of your code
}

answered Feb 15 '14 at 11:41

Drixson Oseña

3,631
3
23
36

@KiranGopalakrishnan add `method="POST"` in your form since your using post value – Drixson Oseña Feb 15 '14 at 12:29

score 1 · Answer 4 · answered Jan 11 '15 at 09:51

1

Try this

    $("#postcontent").submit(function() {
  return false;
};

$('#postsubmit').click(function(){

// your ajax request here
});

answered Jan 11 '15 at 09:51

Ashwin Sinha

85
2
9

score 1 · Answer 5 · answered Apr 11 '16 at 05:59

1

I think you have putted e.preventDefault(); before ajax call that's why its prevent calling of that function and your Ajax call will not call.

So try to remove that e.prevent Default() before Ajax call and add it to the after Ajax call.

answered Apr 11 '16 at 05:59

Pravin Tukadiya

489
4
20

This is what I looked for. Thank you. – namal Jul 01 '18 at 11:52

score 1 · Answer 6 · answered Jul 24 '16 at 23:56

For the sake of documentation. I spent two days working out an ajax problem and this afternoon when I started testing, my PHP ajax handler wasn't getting called....

Extraordinarily frustrating.

The solution to my problem (which might help others) is the priority of the add_action.

add_action ('wp_ajax_(my handler), array('class_name', 'static_function'), 1);

recalling that the default priority = 10

I was getting a return code of zero and none of my code was being called.

...noting that this wasn't a WordPress problem, I probably misspoke on this question. My apologies.

score 0 · Answer 7 · edited Apr 24 '18 at 07:36

Give this a go:

<form name="postcontent" id="postcontent">
    <input name="postsubmit" type="submit" id="postsubmit" value="POST"/>
    <textarea id="postdata" name="postdata" placeholder="What's Up ?"></textarea>
</form>
<script>
    (function() {
        $("#postcontent").on('submit', function(e) {
            e.preventDefault();
        $.ajax({
            type:"POST",
            url:"add_new_post.php",
            data:$("#postcontent").serialize(),
            beforeSend:function(){
                $(".post_submitting").show().html("<center><img src='images/loading.gif'/></center>");
            },success:function(response){   
                //alert(response);
                $("#return_update_msg").html(response); 
                $(".post_submitting").fadeOut(1000);                
            }
        });
   });
})();
</script>

score 0 · Answer 8 · answered Oct 24 '18 at 17:26

I am doing a code like this

<!-- Optional JavaScript -->
<!-- jQuery first, then Popper.js, then Bootstrap JS -->
<script
    src="http://code.jquery.com/jquery-3.3.1.min.js"
    integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
    crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.3/umd/popper.min.js" integrity="sha384-ZMP7rVo3mIykV+2+9J3UJ46jBk0WLaUAdn689aCwoqbBJiSnjAK/l8WvCWPIPm49" crossorigin="anonymous"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.1.3/js/bootstrap.min.js" integrity="sha384-ChfqqxuZUCnJSK3+MXmPNIyE6ZbWh2IMqE241rYiqJxyMiZ6OW/JmZQ5stwEULTy" crossorigin="anonymous"></script>
<script>
    $(document).ready(function () {

        $(".work-category a").click(function (e) {
            e.preventDefault();
            var id = $(this).attr('data-id');
            $.ajax({
                url: 'process.php',
                method: 'POST',
                data: {
                    clickCategoryID : id
                },
                dataType: 'JSON',
                success: function (data) {
                    $("#content-area").html(data.Content);
                    $(".container-area").animate({top: '100px'}, 1000);
                    $(".single-content").animate({opacity:1}, 1000);
                }
            });
        });

    });
</script>

But the code is not running and the console saya process.php not found though I have the code on it.

This suggests that you have `process.php` inside the same folder that the html file is in , Is that the case ? — Kiran Gopalakrishnan, Oct 25 '18 at 13:29

score 0 · Answer 9 · answered Aug 08 '19 at 20:19

I was having this issue and none of these answers were able to help me with the issue. For anyone else who happens to have the same problem, here is a possible solution that worked for me. Simply reformat the code to be the following:

$("#postcontent").submit(function(e) {
  $.ajax({
    type:"POST",
    url:"add_new_post.php",
    data:$("#postcontent").serialize(),
    beforeSend:function(){
      $(".post_submitting").show().html("<center><img src='images/loading.gif'/></center>");
    },
    success (response) {   
      //alert(response);
      $("#return_update_msg").html(response); 
      $(".post_submitting").fadeOut(1000);                
    },
    error (data) {
      // handle the error
    }
  });
});

The difference in the code is that this code has an error handler. You can choose to handle the error however you wish but it is best practice to have the error handler in case of an exception. Also, the formatting of the success and error functions within the ajax function are different. This is the only format for success/error that would work for me. Hope it helps!

score 0 · Answer 10 · answered Dec 07 '19 at 12:47

my html and js code

<script>
$(".editTest23").change(function () {

        var test_date = $(this).data('');id
        // alert(status_id);    

        $.ajax({
            type: "POST",
            url: "Doctor/getTestData",
            data: {
                test_data: test_date,
            },
            dataType: "text",
            success: function (data) {
                $('#prepend_here_test1').html(data);
            }
        });
        // you have missed this bracket
        return false;
    });
</script>

in php code

    foreach($patitent_data as $result){

        $result_html .="<tr class='test_record'>\
        <td><input type='text' name='test_name' value='$result->test_name' class='form-control'></td>\
        <td><textarea class='form-control' name='instruction'> $result->instruction </textarea>\
        </td>\
        <td><button class='close remove_test_record' aria-hidden='true'>&times;</button></td>\
     </tr>";
    }

    echo json_encode($result_html)

score -1 · Answer 11 · answered Aug 14 '23 at 09:13

-1

you can also use Submit if you take input type submit then it will automatically submit the form or data so simply use button tag with its type or use input type button

answered Aug 14 '23 at 09:13

ROHIT KUMAR SINGH

1

3

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Aug 16 '23 at 09:38

jquery ajax function not working

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