webmethod显示500(内部服务器错误)

时间:2014-07-14 05:50:14

标签: javascript asp.net vb.net webmethod

我正在尝试从alertBook.aspx.vb调用webmethod。 但每当我点击按钮时,我收到错误500(内部服务器错误)。

这是从javascript调用webmethod的正确方法吗? 我是javascript的新手。

提前致谢。

我的javascript如下:

function webMethod() {
                setInterval(function () {
                    $.ajax({
                        type: "post",
                        url: "alertBook.aspx/newUid",
                        contentType: "application/json; charset=utf-8",
                        dataType: "json",
                        success: function (result) {
                            OnSuccess(result);
                        },
                        error: function (xhr, status, error) {
                            OnFailure(error);
                        }
                    });
                }, 3000);
            }
            function OnSucceeded(response) {
                alert(response);
                //pushNoti(response, "info", "y");
            }

            function OnFailure(error) {
                alert(error);
            }

alertBook.aspx是

<a class="btn btn-info" onclick="webMethod()">WebMethod</a>

我的网络方法如下:

<WebMethod()> _
    Public Shared Function newUid(ByVal msg As String) As String

        Dim objForm As New alertBook

        Dim objUtl As New uClass.fUtilities

        Dim sqlQuery As String
        Dim UID1, UID2 As String
        Dim notiStr As String
        Dim objDb As New uClass.dbFunctions("myCon")

        Dim dtRecords As DataTable

        sqlQuery = "select top 1 UID from alertsBook order by UID desc"

        If objDb.getDataTable(dtRecords, sqlQuery) = False Then
            Return objForm.displayMessage("Error", objDb.errDesc)
        End If

        For rCount = 0 To dtRecords.Rows.Count - 1
            UID1 = dtRecords.Rows(rCount)("UID")
        Next

        If UID1 <> UID2 Then
            notiStr = "<script> " & displayNoti("", "info", "New Message", True) & "</script>"
            UID2 = UID1
            'scriptDiv.InnerHtml = notiStr
        Else
            notiStr = ""
        End If

        Return notiStr

    End Function

1 个答案:

答案 0 :(得分:1)

您应该查找有关错误的更多详细信息。你可以使用

  error: function (xhr, status, error){
    alert(xhr.status);
    alert(xhr.responseText);
  }

xhr.responseText包含有关ajax错误的详细信息 获取更多细节或错误并相应地更改代码。