jQuery - submit multiple forms through single request, without Ajax

Question

I have a page with several forms. I am trying to submit one of the forms (say form A) (NOT through Ajax, since I need to load the result page after the submit is processed), BUT I need another form's contents (say form B) to be submitted TOGETHER with form A, i.e. the contents of forms A + B should be submitted TOGETHER to the SAME URL, as one request, and as I said before, NOT as an Ajax request.

The submit should be by a POST request. Also, the forms contain ONLY fields that are not file upload (i.e. input, select, textarea fields).

I have seen suggestions here such as

Posting/submitting multiple forms in jQuery

or

Submitting two forms with a single button

but pay attention that these do not match my case, since the forms are submitted in different requests, and/or they are submitted through Ajax.

I was thinking of getting the contents of one of the forms by (the jQuery's) serialize(), but how do I attach that string to a form submitted by POST?

Or maybe you have other ideas how to accomplish this?

SOLUTION:

Based on the ideas of Sheepy, I wrote the following code. I am also using the answer by Pointy from the following link:

submit multiple forms to same page

//Arguments: "name"s of forms to submit.
//First argument: the form which according to its "action" all other forms will be submitted.
//Example: mergeForms("form1","form2","form3","form4")
function mergeForms() {
    var forms = [];
    $.each($.makeArray(arguments), function(index, value) {
        forms[index] = document.forms[value];
    });
    var targetForm = forms[0];
    $.each(forms, function(i, f) {
        if (i != 0) {
            $(f).find('input, select, textarea')
                .hide()
                .appendTo($(targetForm));
        }
    });
    $(targetForm).submit();
}

Sheepy · Accepted Answer · 2011-08-14 15:59:15Z

8

Well, you have to copy the data from form2 to form1 before the submit. Here is the basic to get you started:

$.each ( $('#form2 input, #form2 select, #form2 textarea').serializeArray(), function ( i, obj ) {
  $('<input type="hidden">').prop( obj ).appendTo( $('#form1') );
} );

This function would select inputs from form2, get their current values, then create a new hidden input for each of them and add them to form1.

Depending on your scenario you may want to check the existance of input with same name on form1 first.

answered Aug 14, 2011 at 15:59

Sheepy

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2 Comments

Chris Over a year ago

This won't include disabled inputs

Sheepy Over a year ago

Yes @Chris. Disabled inputs are not submitted normally so it's expected. jQuery's serializeArray respect the disabled property too, you'll need to enable them before the code and re-disable them after.

Julien Lafont · Accepted Answer · 2011-08-14 16:01:05Z

3

A solution to inject the data directly in the post request (and not in a sub field)

<form id="form1">
    <input ... />
    <input ... />
</form>

<form id="form2">
    <input ... />
    <input ... />
</form>

<form id="result" action="..." method="POST" onsubmit="merge(); return true;">
    <input type="submit" />
</form>


<script type="text/javascript">
    function merge() {
        $result = $("#result");
        $("#form1 input, #form2 input, #form1 select, #form2 select, #form1 textarea, #form2 textarea").each(function() {
            $result.append("<input type='hidden' name='"+$(this).attr('name')+"' value='"+$(this).val()+"' />");
        });
    }
</script>

answered Aug 14, 2011 at 16:01

Julien Lafont

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Comments

Stephen · Accepted Answer · 2011-08-14 15:49:52Z

2

You can only do one request at a time and that will reload the page unless you use AJAX. The only other option would be to create a script which concatantes the seralizations of your two forms - but at that point, why wouldn't you just use one large form..

Edit: For your combining the two forms, there is an example of this in the answer of the second link you provided.

answered Aug 14, 2011 at 15:49

Stephen

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2 Comments

rapt Over a year ago

Depending on the scenario, the user might want to submit only form B; or form A (in which case form B also should be submitted).

James Alday Over a year ago

In my case, I have a 'search' form and a 'search filters' form. They are intrinsically tied together but displayed separately in the ui, but combining their values via javascript is probably an easier than trying to figure out the layout to get them both to display without disturbing any other forms the page may have...

Mihai Iorga · Accepted Answer · 2011-08-14 15:51:04Z

1

form A:

<form method="post" action="next.html" onsubmit="this.formbVal.value = $('#formb').serialize(); return true;">
    <input type="hidden" name="formbVal" value="" />
    <input type="submit">
</form>

form B:

<form id="formb" method="post" action="#" onsubmit="return false;">
</form>

answered Aug 14, 2011 at 15:51

Mihai Iorga

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1 Comment

rapt Over a year ago

The problem with this is that I get the whole form B as a single new value. But the code on the server expects a specific list of values that matches form B (sometimes form B is submitted without any other form). I can go ahead and analyze formVal on the server and break it down but it's less elegant. Let the javascript send the form in an acceptable format.

Leigh · Accepted Answer · 2013-03-14 04:46:15Z

I discovered that your solution will copy the form to another invissible form, but the backside is that the original forms will be cleared an such edit of the original form (eg after form validation errors) is not possible.

To solve this i tried to clone the elements before appending it to the invissible form. I edited your code like this:

//Arguments: "name"s of forms to submit.
//First argument: the form which according to its "action" all other forms will be submitted.
//Example: mergeForms("form1","form2","form3","form4")    
function mergeForms() {
        var forms = [];
        $.each($.makeArray(arguments), function(index, value) {
            forms[index] = document.forms[value];
        });
        var targetForm = forms[0];
        $.each(forms, function(i, f) {
            if (i != 0) {
                $(f).find('input, select, textarea')
                    .clone()
                    .hide()
                    .appendTo($(targetForm));
            }
        });

Unfortunatly now it doesn't copy the selected values in a select element.

Hope someone can help me to impove this script

kleinohad · Accepted Answer · 2011-08-14 15:59:29Z

-1

you can do some thing like this

$('#SubmitBtn).click(function () {
        $("#formId").attr("action", "http://www.YourSite.com/");
        $("#formId").attr("method", "post");
        $("#formId").attr("target","_blank");
        $('#formId').submit();
    });

this will submit the for to http://www.YourSite.com/ in a new window

answered Aug 14, 2011 at 15:59

kleinohad

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