dongtuo0828
2019-01-09 16:33 阅读 41
已采纳

如何通过单击按钮在一个表单提交中调用两个操作?

I Have a form in PHP. when I am clicking the submit button I want to take two actions at the same time. how do I do that?

<script>
     function myfunction(){

          $.ajax({
            type: 'post',
            url: 'merchants.php',
            data: $('form').serialize(),
            success: function () {
              alert('form was submitted');
            }
          });
      }


    </script>

<div class="stdFormHeader"> New Merchant Registration</div>
<form action="" method="POST">
    <label class="stdFormLabel">Merchant name : </label><input class="stdFormInput" type="text" name="merchantName" required><br>
<!--    <label class="stdFormLabel">Business Type : </label><select class="stdFormSelect" name="shopMarket" required>-->
<!--        <option value="shop">Shop</option>-->
<!--        <option value="market">Market Place</option>-->
<!--    </select><br>-->
    <label class="stdFormLabel">Contact Person : </label><input class="stdFormInput" type="text" name="contactPerson" required><br>
    <label class="stdFormLabel">Contact Number : </label><input class="stdFormInput" type="text" name="contactNumber" required><br>
    <label class="stdFormLabel">Address : </label><textarea class="stdFormInputBox" name="address"></textarea><br>

    <input class="stdFormButton"  type="submit" name="submit" onclick="myfunction()" value="Apply">
</form>
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1条回答 默认 最新

  • 已采纳
    dongsu2807 dongsu2807 2019-01-09 17:03

    Just do a submit again:

     function myfunction(){
    
      $.ajax({
        type: 'post',
        url: 'merchants.php',
        data: $('form').serialize(),
        success: function () {
          alert('form was submitted');
        }
      });
    
      $.ajax({
        type: 'post',
        url: 'OtherFunction.php',
        data: $('form').serialize(),
        success: function () {
          alert('form was submitted again');
        }
      });
    }
    
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