提交表格并留在同一页面上？

Question

I have a form that looks like this

<form action="receiver.pl" method="post">
  <input name="signed" type="checkbox">
  <input value="Save" type="submit">
</form>

and I would like to stay on the same page, when Submit is clicked, but still have receiver.pl executed.

How should that be done?

Answer 1

The easiest answer: jQuery. Do something like this:

$(document).ready(function(){
   var $form = $('form');
   $form.submit(function(){
      $.post($(this).attr('action'), $(this).serialize(), function(response){
            // do something here on success
      },'json');
      return false;
   });
});

If you want to add content dynamically and still need it to work, and also with more than one form, you can do this:

   $('form').live('submit', function(){
      $.post($(this).attr('action'), $(this).serialize(), function(response){
            // do something here on success
      },'json');
      return false;
   });

Answer 2

99% of the time I would use XMLHttpRequest or fetch for something like this. However, there's an alternative solution which doesn't require javascript...

You could include a hidden iframe on your page and set the target attribute of your form to point to that iframe.

<style>
  .hide { position:absolute; top:-1px; left:-1px; width:1px; height:1px; }
</style>

<iframe name="hiddenFrame" class="hide"></iframe>

<form action="receiver.pl" method="post" target="hiddenFrame">
  <input name="signed" type="checkbox">
  <input value="Save" type="submit">
</form>

There are very few scenarios where I would choose this route. Generally handling it with javascript is better because, with javascript you can...

• gracefully handle errors (eg retry)
• provide UI indicators (eg loading, processing, success, failure)
• run logic before the request is sent, or run logic after the response is received.

Answer 3

HTTP / CGI方法是让程序返回HTTP状态代码204（无内容）。

Answer 4

When you hit on the submit button, the page is sent to the server. If you want to send it async, you can do it with ajax.

Answer 5

Use XMLHttpRequest

var xhr = new XMLHttpRequest();
xhr.open("POST", '/server', true);

//Send the proper header information along with the request
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");

xhr.onreadystatechange = function() { // Call a function when the state changes.
    if (this.readyState === XMLHttpRequest.DONE && this.status === 200) {
        // Request finished. Do processing here.
    }
}
xhr.send("foo=bar&lorem=ipsum");
// xhr.send(new Int8Array()); 
// xhr.send(document);

Answer 6

Just use: action="" . There's no need for anything like javascript.