Question

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>

Answer 1

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。

jquery表单没有提交

1 个答案: