javascript - 如何提交附加表格

Question

每次单击时我都有一个“更多”按钮来克隆表单。我想在单击每个表单的“提交”按钮时使用 ajax 提交每个表单。如何使用 Ajax 方法分别提交每个表单？这是我的代码:

$(document).ready(function() {
  $(".More").click(function() {
    $(".MyForm")
      .eq(0)
      .clone()
      .show()
      .insertAfter(".MyForm:last");
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.1/jquery.min.js"></script>
<span class="More">More+</span>
<div class="all">
  <form class="MyForm" method="post">
    <input type="text" placeholder="name" value="Aynaz" />
    <select name="Avg">
      <option value="1">1</option>
      <option value="2">2</option>
    </select>
    <button type="submit">Submit</button>
  </form>
</div>

Answer 1

在您拥有的每个 submit 按钮上添加一个 click 监听器，停止默认操作(以确保未提交特定表单)，并创建新的每种形式的ajax:

$(document).ready(function() {
  $(".More").click(function() {
    $(".MyForm")
    .eq(0)
    .clone()
    .show()
    .insertAfter(".MyForm:last");
  });

  $(document).on('click', '.MyForm button[type=submit]', function(e) {
    e.preventDefault() // To make sure the form is not submitted 
    $('.MyForm').each(function() {
      console.log($(this).serialize())
      $.ajax(
        $(this).attr('action'), 
        {
          method: $(this).attr('method'),
          data: $(this).serialize()
        }
      )
    });
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.1/jquery.min.js"></script>
<span class="More">More+</span>
<div class="all">
  <form class="MyForm" method="post">
    <input type="text" placeholder="name" value="Aynaz" name="a1" />
    <select name="Avg">
      <option value="1">1</option>
      <option value="2">2</option>
    </select>
    <button type="submit">Submit</button>
  </form>
</div>

Note - i had to add a name attribute to your text input (inputs without names are not being submitted).