Upload file via ajax jquery php api

Question

I am having the code with the below format.

PHP file :

<form action="http://clientwebapi.com/createEvent" id="form_createEvent" method="post" enctype="multipart/form-data">
<input type="text" name="image_title" />
<input type="file" name="media" accept="image/*,video/*"/>
</form>

JQUERY file:

$('#form_createEvent').submit(function () {
    var form = $(this);
    $.ajax({
        url: form.attr("action"),
        type: form.attr("method"),
        xhrFields: {
            withCredentials: true
        },
        data: form.serialize()
    }).done(function () {
        showCurrentLocation();
        alert('Event created successfully..');
        location.reload();

    }).fail(function () {
        alert("fail!");
    });
    event.preventDefault();
});

The above Jquery code is submitting. Also While I am submitting the below format, It will redirect to the "http://clientwebapi.com/createEvent" and Event created successfully.

Form Submit and redirect to client page:

$('#form_createEvent').submit(function () {
    var fd = new FormData();
    fd.append('media', input.files[0]);

    $.ajax({
        url: form.attr("action"),
        data: fd,
        processData: false,
        contentType: false,
        type: form.attr("method"),
        success: function (data) {
            alert(data);
        }
    });
event.preventDefault();

});

Here How can I prevent while submit the form and create the event with the given image. Kindly help

Actually I am sending the datas including image to "clientwebapi.com/createEvent". After processing this, it will send some response. I need to know how can we pass image to client API via this jquery, ajax code — Arun Rama Balan.G
– Arun Rama Balan.G, Commented Aug 28, 2013 at 10:38

Michiel · Accepted Answer · 2013-08-28 09:54:10Z

1

You forgot to add event as argument to the submit function:

$('#form_createEvent').submit(function (event) {

answered Aug 28, 2013 at 9:54

Michiel

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1 Comment

Ben Fortune Over a year ago

That's because you have event.preventDefault(); after your ajax code.

Arun Rama Balan.G · Accepted Answer · 2013-08-30 10:48:01Z

I found the Answer for this. I made some mistake here. I resolved by using the below code..

$('#form_createEvent').submit(function() { 
            var form = new FormData($(this)[0]); 
            $.ajax({
                url: 'http://clientwebapi.com/createEvent/',
                type: 'POST',
                xhrFields: {
                    withCredentials: true
                },
                async: false,
                cache: false,
                contentType: false,
                processData: false,
                data: form,
                beforeSend: function () {
                    $("#output_process").html("Uploading, please wait....");
                },
                success: function () { 
                    $("#output_process").html("Upload success.");
                },
                complete: function () {
                    $("#output_process").html("upload complete.");
                },
                error: function () {
                    //alert("ERROR in upload");
                    location.reload();
                }
            }).done(function() { 
                alert('Event created successfully..');

            }).fail(function() {
                alert("fail!");
            });
            event.preventDefault();
        });

zzlalani · Accepted Answer · 2013-08-28 09:57:17Z

0

you can stop the form submission by return false;

$('#form_createEvent').submit(function () {
    // some code
    $.ajax({
        // some code/objects
    });

    return false; // here, it will stop the form to be post to the url/action

});

answered Aug 28, 2013 at 9:57

zzlalani

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2 Comments

zzlalani Over a year ago

Sorry @GARB i didn't notice that you are trying to upload file, what you actually need is to check this malsup.com/jquery/form

zzlalani Over a year ago

either use this code malsup.com/jquery/form/#ajaxForm or this code malsup.com/jquery/form/#ajaxSubmit directly into your code and see the magic.

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