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jQuery force form POST

trevor1940
trevor1940 asked
on
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Last Modified: 2017-03-08
Hi
I have dynamically generated page with multiple forms on. When a form is submitted I wish to show a simple success / fail message within the form wrapper.

If I disable the javascript the form is received by the back end PHP via $_POST but with the javascript enabled it's received via $_GET!

How do I force the jQuery to 'post' the form?

<!doctype html>
<html>
<head>
<title>Test form</title>
<link rel="stylesheet" href="/jquery-ui-1.9.2.custom/development-bundle/themes/base/jquery.ui.all.css">
<script type="text/javascript" src="/jquery-ui-1.9.2.custom/js/jquery-1.8.3.js"></script>
<script type="text/javascript" src="/jquery-ui-1.9.2.custom/js/jquery-ui-1.9.2.custom.min.js"></script>
<script>
$(document).on("submit","form", function(evt){
  evt.preventDefault() ; // stop the form submitting
// alert($(this).attr("action") + " " + $(this).serialize() )
//Find the forms wrapper get the form's action submit it & load the responce into the forms wrapper
  $(this).closest(".FormWrap").load($(this).attr("action") + "?" + $(this).serialize() )
})

</script>

<style type="text/css">
body {
font-family:Arial;
font-size:12px;
background:#ededed;
}

#PageWrap{
width: 1000px;
margin: auto;
}

</style>
</head>
  <body>
    <div id="PageWrap">
      <header><img src="/images/Banner.jpg" > </header>
      <div class="FormWrap">
        <fieldset>
          <legend>My Form
          <legend>
          <form id="MyFrm_1"class="form" action="SubmitForm.php" method="post">
            <label for="user_1" >Name</label><input id="user_1" type="text" name="user"value ="Holly">
            <label for="age_1" >Age</label><input type="text" id="age_1" name="age" value ="23">
            <input type="submit" name="submit" value="submit" />
          </form>
        </fieldset>
      </div>
    </div>
</body>
</html>

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SubmitForm.php
<?php
if(isset($_POST['Name'])){
  echo "post Name " . $_POST['Name'];
}
elseif(isset($_GET['Name'])){
  echo "GET Name " . $_GET['Name'];
}
else{
  echo "Wot no form";
}

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Comment
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Banshi lal dangiFull Stack Developer

Commented:
jQuery.load() (only does GET unless you pass it an object for data, then POST)

Using, jQuery.ajax(), you can make it POST

$.ajax({
    'url': 'postTo.php',
    'type': 'POST',
    'data': $('#yourFormId').serialize(),
    'success': function(result){
         //process here
    }
});

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Author of the Year 2014

Commented:
Here's the entire "hello world" example using jQuery and PHP.  It shows how to do the request and response into the same page.
https://www.experts-exchange.com/articles/10712/The-Hello-World-Exercise-with-jQuery-and-PHP.html

Author

Commented:
@Julian
That worked thanx
Can you tell me why you use different 'this'

   url: this.action,
   data: $(this).serialize(),

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Most Valuable Expert 2017
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Commented:
Can you tell me why you use different 'this'
Good question!

The $.ajax() call is a jQuery function. When you select elements using jQuery Ex $('form') what you are doing is calling a jQuery function $() and passing it a string or object that is used to find the element or elements you are wanting to target. jQuery finds those elements and wraps them in a jQuery object making the jQuery library accessible on the return from that function.

From that object to get to an attribute one needs to use the .attr() method. To get the action attribute using jQuery we would have to do this
$(this).attr('action');

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But ... in the event handler this refers to the DOM Node - not a jQuery wrapped DOM Node - so you have access to all the default methods and properties you would in normal JavaScript. In normal JavaScript you can get the action as a property on the form element - so it makes no sense to go wrapping the this in a jQuery object to get access to the action attribute when it is natively available on the this in the event handler.
EDIT
In the second this
$(this).serialize()

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We are accessing a jQuery function - so we have to wrap the form element in a jQuery object to be able to use that function.

Author

Commented:
Thank you for solution and additional explanation
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Commented:
You are welcome.