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jquery表单没有提交

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这是我写的用于验证后提交表单的脚本,并使用$ _POST在php文件中接收值,但我没有在php文件中获取值 . 当我试图回应那些在php中显示空白的值 . 请指导我,我是jquery的新手

<script>
$("#changepassform").validate({


  rules: {
    old_password: "required",
    password: "required",
    password2: {
      equalTo: "#password"
    },
  },
  messages: {

    old_password: "Please enter old password",

    password: "Please enter new password",
    password2: " Enter Confirm Password Same as Password"

  },
  submitHandler: function(form) {

      var current_password = $("#current_password").val();
      var new_password = $("#password").val();
      var comfirm_password = $("#password2").val();
      var id = $("#id").val();
      var dataString = 'newpassword1=' + new_password + '&id1=' + id;

      $.ajax({
        type: "POST",
        url: "changepassword.php",
        data: "dataString",
        success: function(response) {

          $("#status").html(response);
        }
      });

    }
    //form.submit();

  //return false;  

});
// required to block normal submit since you used ajax
//form.submit();
</script>

1 回答

  • 0

    发送数据时出错

    <script>
        $("#changepassform").validate({
    
    
          rules: {
            old_password: "required",
            password: "required",
            password2: {
              equalTo: "#password"
            },
          },
          messages: {
    
            old_password: "Please enter old password",
    
            password: "Please enter new password",
            password2: " Enter Confirm Password Same as Password"
    
          },
          submitHandler: function(form) {
    
              var current_password = $("#current_password").val();
              var new_password = $("#password").val();
              var comfirm_password = $("#password2").val();
              var id = $("#id").val();
              var dataString = 'newpassword1=' + new_password + '&id1=' + id;
    
              $.ajax({
                type: "POST",
                url: "changepassword.php",
                data: dataString,
                success: function(response) {
    
                  $("#status").html(response);
                }
              });
    
            }
            //form.submit();
    
          //return false;  
    
        });
        // required to block normal submit since you used ajax
        //form.submit();
        </script>
    

    使用 serialize() 方法 . 在发出AJAX请求时,可以在URL查询字符串中使用序列化值 .

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