Question

17

How do I process a < div> as a form < input>

rated 0 times [ 23] [ 6] / answers: 1 / hits: 44278 / 11 Years ago, thu, march 21, 2013, 12:00:00

What I've done is:

HTML

<form>

    <div id=textBox contenteditable=true name=textBox><?php echo $storyText; ?>

    </div>

    <textarea id=hiddeninput name=hiddeninput></textarea>

    <input type=submit id=save name=save value=Submit/>

</form>

Javascript

$('#save').click(function () {

    var mysave = $('#textBox').html();

    $('#hiddeninput').val(mysave);

    $(form:first).submit();

    $('#hiddeninput').append(mysave);

    alert($('#hiddeninput').val());

});

So both the alert and the append display the correct information, but it won't save #hiddeninput as a php variable when I submit. Originally I had this as an hidden input method, but I'm trying to show that it won't post no matter what I do,

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raymondd

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answered 11 Years ago alyssiat · Accepted Answer

Your code is working almost as it is.
But I'd rather use normal <input type=hidden> and you don't need to trigger submit for your form in your case just put the value in a hidden field.

Given your markup, with slight modifications

<form action=showrequest.php>

<div id=textBox contenteditable=true name=textBox style=width:300px;height:100px;>

</div>

<textarea id=hiddeninput name=hiddeninput></textarea>

<input type=submit id=save name=save value=Submit/>

</form>

js

$(function(){

    $('#save').click(function () {

        var mysave = $('#textBox').html();

        $('#hiddeninput').val(mysave);

    });

});

var_dump($_REQUEST) on php side gives

array(2) {

  [hiddeninput]=>

  string(4) test

  [save]=>

  string(6) Submit

}