PHP submit Multiple Forms with 1 button and 1 action

Question

I want to submit Multiple Forms with 1 button and 1 action target using PHP. Is is possible ?

HTML

<form name="myform" action="test_post.php" method="post">
Name: <input type='text' name='name' />
</form>

<form name="myform2" action="test_post.php" method="post">
Class: <input type='text' name='class' />
</form>

<a href="javascript: submitform()">Search</a>

JS

function submitform()
{
document.myform.submit();
document.myform2.submit();
}

PHP (test_post.php)

echo $name = $_POST['name'];
echo $class = $_POST['class'];

I tried with that code but it just show $_POST['class'] value. For name it show error : Undefined index: name in...

Please advice.

Answer 1

you dont need one form per input, you can have a million in one one form so ..

<form name="myform" action="test_post.php" method="post">
Name: <input type='text' name='name' />

Class: <input type='text' name='class' />
</form>

<a href="javascript: submitform()">Search</a>

should be fine, and you don't really need js to submit. a html submit input is better supported

<input id="submit" type="submit" value="submit">

Answer 2

You actually want two fields on one form - which you can then submit with no problems:

<form name="myform" action="test_post.php" method="post">
Name: <input type='text' name='name' />
Class: <input type='text' name='class' />
<input type='submit'>
</form>

Or you can submit it by using some JS anyhow if you do other thigns in the JS code.

Answer 3

if jquery is an option, then .deferred might be what you are looking for.

function submitform(){
//define a variable where we will store deferred objects
var def = true;

$("form").each(function() {

    var postResult = $.Deferred();

    //.when takes a deferred object as a param. 
    // if we don't pass a deferred object .when treats it as resolved.
    // as our initial value of def=true, the first .when starts immediately
    $.when(def).then(function(){
        $.post('post_destination.php', form.serialize()).done(function(){
            //the chain will fail after the first failed post request
            //if you want all the requests to complete in any case change the above .done to .always
            post.resolve();
        });
    });

    // now we reassign def with the deferred object for the next post request
    def = postResult;
});
}

and here's the link to when I asked the question some time ago. How to submit multiple jquery posts to a page one after another if database updates successfully?