语法错误：JSON.parse：在JSON数据线1列2意外的字符 [英] SyntaxError: JSON.parse: unexpected character at line 1 column 2 of the JSON data

问题描述

控制器

[HttpGet]
public ActionResult VerifyUserEmail(string User_Email)
{
  try
  {
    using (EmptyMVCApplicationEntities objConnection = new EmptyMVCApplicationEntities())
    {
      ObjectParameter objIErrorCode = new ObjectParameter("ErrorCode", typeof(Int32));  
      ObjectParameter objBFlag = new ObjectParameter("bFlg", typeof(bool));
      objConnection.Check_User_Exists(User_Email, objBFlag, objIErrorCode);
      if (Convert.ToBoolean(objBFlag.Value) != true)
      {
        return Json(new { Success = "false", Message = "Email exists" }, JsonRequestBehavior.AllowGet);
      }
      else
      {
        return Json(new { Success = "True", Message = "Email not exists" }, JsonRequestBehavior.AllowGet);
      }               
    }
  }
  catch (Exception Ex) 
  {    
  }
}

剧本

$("#User_Email").blur(function () {
  if ($(this).val() != "") {
    $.ajax({
      url: "/User/VerifyUserEmail?User_Email=" + $("#User_Email").val(),
      success: function (result) {
        try {
          var jsonIssueObj = $.parseJSON(result).Data;    
        }  catch (e) { alert(e); }
        if (!jsonIssueObj.Success) {
          var errorMsg = jsonIssueObj.Message;
          $('#msg').html(errorMsg);
          $('#msg').show();
        }
        else {
          var errorMsg = null;
          $('#msg').html(errorMsg);
          $('#msg').hide();
        }
      }
    });
  }
  return false;
});

我收到以下错误：

I am getting below error:

语法错误：JSON.parse：第1行的JSON数据第2栏意外的字符

SyntaxError: JSON.parse: unexpected character at line 1 column 2 of the JSON data

另外，如果我想成功和消息中的对象，如何传递一个对象作为控制器的ActionResult JSON。

Also If I want Success and Message in an object, how to pass an object as json in controller actionresult.

推荐答案

您在返回与2个属性JS​​ON，成功和消息（你不需要 parseJSON ）。要访问它们。

You returning json with 2 properties, Success and Message (you do not need parseJSON). To access them

success: function (result) {
  var success = result.Success;
  var message = result.Message;
  ....

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