使用JavaScript手动提交表单不会发送提交按钮

Question

I have a form that has two submit buttons. I want to submit the form manually using JavaScript and have the input button used to submit the form posted along with the other form elements, as it would be if the form was submitted automatically. There's quite a lot of chatter on this subject, but I can't find an answer.

<form method="post" action="echoToScreenAndLog.jsp" id="form1"> 
    <input id="field1" name="field1"/>
    <input type="text" size="20" id="field2" name="field2"/>

    <input type="submit" value="Do One" name="sub1_name" id="sub1_id"/>
    <input type="submit" value="Do Two" name="sub2_name" id="sub2_id"/>
</form> 

When the form is submitted above using the "Do One" button, the posted parameters are field1="xxx" , field2="yyy" , sub1_name="Do One" .

But I want to submit the form manually...

<form method="post" action="echoToScreenAndLog.jsp" id="form1"> 
    <input id="field1" name="field1"/>
    <input type="text" size="20" id="field2" name="field2"/>

    <input type="submit" value="Do One" name="sub1_name" id="sub1_id"/>
    <input type="submit" value="Do Two" name="sub2_name" id="sub2_id"/>
</form> 
<script type="text/javascript">
    var btn = document.getElementById('sub1_id');
    btn.onclick=function() {
        return mySubmit(document.getElementById('form1'), ...);
    }
</script>

but doing a manual submission of the form in the mySubmit function does not post the sub1_name parameter. I can understand that - I've bypassed the submission so the form is not being submitted using the buttons and therefore it makes no sense to post a parameter representing the button used to submit the form.

When I look at the elements of the form in the onclick handler, I can see both buttons. I'm not overly surprised by that either, they are elements on the form after all, but what I don't get is that if I add an element inside my onclick handler then the element I add IS posted and the two original submit buttons are not posted. Just to complete the picture, here's the code that adds the element:

<script type="text/javascript">
    var btn = document.getElementById('sub1_id');
    btn.onclick=function() {
        var f = document.getElementById('form1');
        var s = document.createElement("input");
        s.type="hidden"; s.name="xsubmit_name"; s.value="Bob"; s.id="xsubmit_id";            
        f.appendChild(s);

        // s gets posted
        return mySubmit(f, ...);
    }
</script>

Adding the input element could work for me, but I'm confused how the browser knows to post my added element and not the original two input elements.

Thank you.

Answer 1

The specification says that the first step for form submission is:

Step one: Identify the successful controls

"Successful controls" are defined as:

A successful control is "valid" for submission. Every successful control has its control name paired with its current value as part of the submitted form data set. A successful control must be defined within a FORM element and must have a control name.

However:

...

• If a form contains more than one submit button, only the activated submit button is successful.

Since none of the submit buttons are activated, none are sent. Hidden input elements, on the other hand, are valid and will just be submitted along. Note that you add the hidden elements before calling mySubmit() , so at the time the above steps are executed (ie during submit), the hidden element is just another successful control part of the form, and thus sent.

Answer 2

may use

var btn = document.getElementById('sub1_id');
btn.onsubmit=function() {
   return false;
}

btn.onclick=function() {
    var f = document.getElementById('form1');
    var s = document.createElement("input");
    s.type="hidden"; s.name="xsubmit_name"; s.value="Bob"; s.id="xsubmit_id";            
    f.appendChild(s);

    f.submit()
}