CODEWITHSUNDEEP
phpjqueryformsserializationform-submit
Dan Dinu
Dan Dinu

Reputation: 33368

Serializing and submitting a form with jQuery and PHP

I'm trying to send a form's data using jQuery. However, data does not reach the server. Can you please tell me what I'm doing wrong?

My HTML form:

<form id="contactForm" name="contactForm" method="post">
    <input type="text" name="nume" size="40" placeholder="Nume">
    <input type="text" name="telefon" size="40" placeholder="Telefon">
    <input type="text" name="email" size="40" placeholder="Email">
    <textarea name="comentarii" cols="36" rows="5" placeholder="Message"></textarea> 
    <input id="submitBtn" type="submit" name="submit" value="Trimite">
</form>


JavaScript (in the same file as the above form): 

<script type="text/javascript">
    $(document).ready(function(e) {

        $("#contactForm").submit(function() {
            $.post("getcontact.php", $("#contactForm").serialize())
            // Serialization looks good: name=textInNameInput&&telefon=textInPhoneInput etc
            .done(function(data) {
                if (data.trim().length > 0) {
                    $("#sent").text("Error");   
                } else {
                    $("#sent").text("Success");
                }
            });

            return false;
        })
    });
</script>


Server side PHP (/getcontact.php):

$nume = $_REQUEST["nume"]; // $nume contains no data. Also tried $_POST
$email = $_REQUEST["email"];
$telefon = $_REQUEST["telefon"];
$comentarii = $_REQUEST["comentarii"];


Can you please tell me what I am doing wrong?

UPDATE

Checked var_dump($_POST) and it returned an empty array.

The weird thing is that the same code tested on my local machine works fine. If I upload the files on my hosting space it stops working. I tried doing an old-fashioned form without using jQuery and all data was correct.

I don't see how this would be a server configuration problem. Any ideas?

Thank you!

Upvotes: 53

Views: 315953

Answers (8)

Hicham O&#39;Sfh
Hicham O&#39;Sfh

Reputation: 841

Here is a more detailed and explained version of @rahulbhondave answer

$('#my-button').click(function () {
    $.post(this.form.action, $(this.form).serializeArray(),
            function (successMessage) {
                console.log(
                    successMessage
                );
            },
            "json"
        )
        .done(function () {
            alert("second success");
        })
        .fail(function () {
            alert("error");
        })
        .always(function () {
            alert("finished");
        });
});

Upvotes: 0

zamil
zamil

Reputation: 1930

You can use this function 

var datastring = $("#contactForm").serialize();
$.ajax({
    type: "POST",
    url: "your url.php",
    data: datastring,
    dataType: "json",
    success: function(data) {
        //var obj = jQuery.parseJSON(data); if the dataType is not specified as json uncomment this
        // do what ever you want with the server response
    },
    error: function() {
        alert('error handling here');
    }
});

return type is json

EDIT: I use event.preventDefault to prevent the browser getting submitted in such scenarios.

Adding more data to the answer.

dataType: "jsonp" if it is a cross-domain call.

beforeSend: // this is a pre-request call back function

complete: // a function to be called after the request ends.so code that has to be executed regardless of success or error can go here

async: // by default, all requests are sent asynchronously

cache: // by default true. If set to false, it will force requested pages not to be cached by the browser.

Find the official page here

Upvotes: 110

Đọc truyện hay
Đọc truyện hay

Reputation: 2023

You can add extra data with form data

use serializeArray and add the additional data:

var data = $('#myForm').serializeArray();
    data.push({name: 'tienn2t', value: 'love'});
    $.ajax({
      type: "POST",
      url: "your url.php",
      data: data,
      dataType: "json",
      success: function(data) {
          //var obj = jQuery.parseJSON(data); if the dataType is not     specified as json uncomment this
        // do what ever you want with the server response
     },
    error: function() {
        alert('error handing here');
    }
});

Upvotes: 39

Zigri2612
Zigri2612

Reputation: 2310

Two End Registration or Users Automatically Registered to "Shouttoday" by ajax when they Complete Registration by form submission.

var deffered = $.ajax({
     url:"code.shouttoday.com/ajax_registration",
     type:"POST",
     data: $('form').serialize()
    }); 

        $(function(){ 
            var form;
            $('form').submit( function(event) {
                var formId = $(this).attr("id");
                    form = this;
                event.preventDefault();
                deffered.done(function(response){
                    alert($('form').serializeArray());alert(response);
                    alert("success");
                    alert('Submitting');
                    form.submit();
                })
                .error(function(){
                    alert(JSON.stringify($('form').serializeArray()).toString());
                    alert("error");
                    form.submit();
                });
            });
         });

Upvotes: 0

rahulbhondave
rahulbhondave

Reputation: 186

 $("#contactForm").submit(function() {

    $.post(url, $.param($(this).serializeArray()), function(data) {

    });
 });

Upvotes: 1

DXB-DEV
DXB-DEV

Reputation: 579

The problem can be PHP configuration:

Please check the setting max_input_vars in the php.ini file.

Try to increase the value of this setting to 5000 as example.

max_input_vars = 5000

Then restart your web-server and try.

Upvotes: 0

A. M. M&#233;rida
A. M. M&#233;rida

Reputation: 2618

Have you looked in firebug if POST or GET?.

check the console display.

Put in the test script:

console.log(data);

You can see the response from the server, if it shows something.

Upvotes: 0

Jan.J
Jan.J

Reputation: 3080

Have you checked in console if data from form is properly serialized? Is ajax request successful? Also you didn't close placeholder quote in, which can cause some problems:

 <textarea name="comentarii" cols="36" rows="5" placeholder="Message>

Upvotes: 0

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