使用jquery-ajax在多个表单中提交表单

i have multiple forms with same class name

我有多个具有相同类名的表单

<form >
  <input type="hidden" value="<%=ids%>" name="idd">
  <input type="hidden" value="<%=email%>" name="cby">
  <input type="text" class="cmd" name="cm" style="width:300px;" placeholder="comment">
  <input type="submit" value="" style="display:none;">
</form>
  <!-- n number of forms are  generated using while loop-->
<form>
  <input type="hidden" value="<%=ids%>" name="idd">
  <input type="hidden" value="<%=email%>" name="cby">
  <input type="text" class="cmd" name="cm" style="width:300px;" placeholder="comment">
  <input type="submit" value="" style="display:none;">
</form>

Then how can i submit a single form among this n number of forms i tried using

那么我怎样才能在我试过的n种表格中提交单一表格

   $(function () {
          $('form').on('submit', function (e) {
                $.ajax({
                 type: 'post',
                  url: 'addfr.jsp',
                  data: $('form').serialize(),
                  success: function () {
                  location.reload();

                  }
          });
         e.preventDefault();
     });
  });

But it is submitting always 1st form among the n-forms. How can submit a random form among the n-forms. May any one help me please.

但它总是在n-forms中提交第一种形式。如何在n-forms中提交随机表格。愿任何人帮助我。

3 个解决方案

#1

You need to reference the form with this when you serialize it...

您序列化时需要引用表单...

$(function () {
    $('form').on('submit', function (e) {
        $.ajax({
            type: 'post',
            url: 'addfr.jsp',
            data: $(this).serialize(),
            success: function () {
                location.reload();
            }
        });
        e.preventDefault();
    });
});

#2

Use this keyword. It will submit the form on which the event has been triggered.

使用此关键字。它将提交触发事件的表单。

why are u using location.reload?

你为什么使用location.reload？

$('form').on('submit', function (e) {
    $.ajax({
    type: 'post',
    url: 'addfr.jsp',
    data: $(this).serialize(),
    success: function () {
        location.reload();
    }
 });

#3

 You can try this...

 function submitForm(){
document.formName.action="actionName";
document.formName.submit();
 }


 <form method="post" name="formName" enctype="multipart/form-data">
  ....
 <input type="button" onclick="submitForm()"/>
 <form>

#1