我有一个.submit()事件设置表单提交。我在页面上也有多个表单,但对于这个示例,这里只有一个表单。我想知道哪个提交按钮在没有应用.click()事件的情况下被单击。

下面是设置:

<html>
<head>
  <title>jQuery research: forms</title>
  <script type='text/javascript' src='../jquery-1.5.2.min.js'></script>
  <script type='text/javascript' language='javascript'>
      $(document).ready(function(){
          $('form[name="testform"]').submit( function(event){ process_form_submission(event); } );
      });
      function process_form_submission( event ) {
          event.preventDefault();
          //var target = $(event.target);
          var me = event.currentTarget;
          var data = me.data.value;
          var which_button = '?';       // <-- this is what I want to know
          alert( 'data: ' + data + ', button: ' + which_button );
      }
  </script>
</head>
<body>
<h2>Here's my form:</h2>
<form action='nothing' method='post' name='testform'>
  <input type='hidden' name='data' value='blahdatayadda' />
  <input type='submit' name='name1' value='value1' />
  <input type='submit' name='name2' value='value2' />
</form>
</body>
</html>

jsfiddle的实例

除了在每个按钮上应用.click()事件外,有没有办法确定哪个提交按钮被单击了?


当前回答

这是我使用的解决方案,效果非常好:

// prevent enter key on some elements to prevent to submit the form function stopRKey(evt) { evt = (evt) ? evt : ((event) ? event : null); var node = (evt.target) ? evt.target : ((evt.srcElement) ? evt.srcElement : null); var alloved_enter_on_type = ['textarea']; if ((evt.keyCode == 13) && ((node.id == "") || ($.inArray(node.type, alloved_enter_on_type) < 0))) { return false; } } $(document).ready(function() { document.onkeypress = stopRKey; // catch the id of submit button and store-it to the form $("form").each(function() { var that = $(this); // define context and reference /* for each of the submit-inputs - in each of the forms on the page - assign click and keypress event */ $("input:submit,button", that).bind("click keypress", function(e) { // store the id of the submit-input on it's enclosing form that.data("callerid", this.id); }); }); $("#form1").submit(function(e) { var origin_id = $(e.target).data("callerid"); alert(origin_id); e.preventDefault(); }); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <form id="form1" name="form1" action="" method="post"> <input type="text" name="text1" /> <input type="submit" id="button1" value="Submit1" name="button1" /> <button type="submit" id="button2" name="button2"> Submit2 </button> <input type="submit" id="button3" value="Submit3" name="button3" /> </form>

其他回答

我问了同样的问题:如何从表单提交事件中获得引起提交的按钮?

我最终想出了这个解决方案,而且效果很好:

$(document).ready(function() {
    $("form").submit(function() { 
        var val = $("input[type=submit][clicked=true]").val();
        // DO WORK
    });
    $("form input[type=submit]").click(function() {
        $("input[type=submit]", $(this).parents("form")).removeAttr("clicked");
        $(this).attr("clicked", "true");
    });
});

在你有多个表单的情况下,你可能需要稍微调整一下,但它仍然适用

对我来说,最好的解决方案是:

$(form).submit(function(e){

   // Get the button that was clicked       
   var submit = $(this.id).context.activeElement;

   // You can get its name like this
   alert(submit.name)

   // You can get its attributes like this too
   alert($(submit).attr('class'))

});

试试这个:

$(document).ready(function(){
    
    $('form[name="testform"]').submit( function(event){
      
        // This is the ID of the clicked button
        var clicked_button_id = event.originalEvent.submitter.id; 
        
    });
});

假设我有这些“提交”按钮:

<button type="submit" name="submitButton" id="update" value="UpdateRecord" class="btn btn-primary">Update Record</button>
<button type="submit" name="submitButton" id="review_info" value="ReviewInfo" class="btn btn-warning sme_only">Review Info</button>
<button type="submit" name="submitButton" id="need_more_info" value="NeedMoreInfo" class="btn btn-warning sme_only">Need More Info</button>

这个"submit"事件处理程序:

 $('#my_form').on('submit', function (e) {
     let x1 = $(this).find("input[type=submit]:focus");
     let x2 = e.originalEvent.submitter.textContent;

两种表达都可以。如果我点击第一个按钮,“x1”和“x2”都返回更新记录。

使用这个出色的答案,您可以检查活动元素(按钮),将隐藏输入附加到表单,并可选地在提交处理程序的末尾删除它。

$('form.form-js').submit(function(event){
    var frm = $(this);
    var btn = $(document.activeElement);
    if(
        btn.length &&
        frm.has(btn) &&
        btn.is('button[type="submit"], input[type="submit"], input[type="image"]') &&
        btn.is('[name]')
    ){
        frm.append('<input type="hidden" id="form-js-temp" name="' + btn.attr('name') + '" value="' + btn.val() + '">');
    }

    // Handle the form submit here

    $('#form-js-temp').remove();
});

附注:我个人在所有通过JavaScript提交的表单上添加了form-js类。