Question

我在ASP.Net web项目的js文件中有以下方法。这种方法给出了错误无法调用null的方法'split'

 function loadUser_InfoCallBack(res)
    {
    var ret=res.value;
    var senderGUID = ret.split(';')[0];
    var senderText = ret.split(';')[1];

    if(senderGUID.Length>0)
    {
    $('hiddenSender.value')=senderGUID;
    $('hiddenText.value')=senderText;
    }

}

任何人都可以提出这个错误的原因..

其中getOwnerInfo方法的代码在

之下

[AjaxPro.AjaxMethod(AjaxPro.HttpSessionStateRequirement.ReadWrite)]
        public string getOwnerInfo()
        {
            string strReturn = ";";

            if (Session["User_GUID"] != null)
            {
                string strUserGUID = Session["User_GUID"].ToString();

                string strIndGUID = "";
                string strIndName = "";

                //Initialize Connection String
                cns = DAL360.Common.getConnection(HttpContext.Current.Session["Server"].ToString(), HttpContext.Current.Session["Database"].ToString());

                DAL360.std_UserRepositoryArtifacts.std_UserRepository userRep = new DAL360.std_UserRepositoryArtifacts.std_UserRepository(cns);
                std_User user = userRep.Getstd_UserByusr_GUID(new Guid(strUserGUID));


                DAL360.std_IndividualRepositoryArtifacts.std_IndividualRepository indRep = new DAL360.std_IndividualRepositoryArtifacts.std_IndividualRepository(cns);
                std_Individual ind = indRep.Getstd_IndividualByind_GUID(new Guid(user.usr_ind_GUID.ToString()));

                if (ind != null)
                {
                    strIndGUID = ind.ind_GUID.ToString(); 

                    strIndName = ind.ind_Prefix + " " + ind.ind_FirstName + " " + ind.ind_MiddleName + " " + ind.ind_LastName;

                    DAL360.std_IndividualAddressRepositoryArtifacts.std_IndividualAddressRepository iadRep = new DAL360.std_IndividualAddressRepositoryArtifacts.std_IndividualAddressRepository(cns);
                    List<std_IndividualAddress> iadList = iadRep.GetAllFromstd_IndividualAddressByIndGUID(ind.ind_GUID);

                    if (iadList != null && iadList.Count > 0)
                    {
                        std_IndividualAddress iad = iadList[0];
                        strIndName += " (" + iad.iad_City + ", " + iad.iad_State + ")";
                    }
                }


                strReturn = strIndGUID + ";" + strIndName;

            }//end if statement


            return strReturn;

        }

此方法位于frmHome.aspx页面的代码隐藏页面中。当我尝试打开页面frmHome.aspx

时，我收到错误

Answer 1

最可能的原因是没有任何元素资源。因此，取res.value将返回null，而split则需要一个字符串。

JS错误：无法调用null方法

1 个答案: