1

很抱歉再次打扰您。

在我们的调查应用程序上，当用户登录并参加我们的一项调查时，下面的代码应该根据参加调查的用户和调查 ID 对调查进行评分。

这些被声明为会话变量。

但是，由于某种原因，该代码无法识别调查 ID 或参与此调查的用户的用户名。

结果，它为每个用户显示重复的答案。

我正在尝试发布屏幕截图，但我遇到了困难。

有什么想法我可能做错了吗？

'//first the surveyId and username are declared in pageLoad event:

        If Session("UserName") Is Nothing Or Session("Username") = "" Then
            Response.Redirect("~/Login.aspx?redirect=List.aspx")
        Else
            userLB.Text = "You are logged in as " & Session("FullName")
            userLB.ForeColor = System.Drawing.Color.DarkOrange
        End If

'//然​​后gridview中的代码：

         <asp:SqlDataSource ID="SqlDataSource1" runat="server" ConnectionString="<%$ ConnectionStrings:DBConnectionString %>"
            SelectCommand="WITH CorrectChoices AS
(
  SELECT Distinct QuestionID, LEFT(CorrectChoice , LEN(CorrectChoice )-1) AS Choice
  FROM SurveyChoices AS extern
  CROSS APPLY
  (
    SELECT Choice + ', '
    FROM SurveyChoices AS intern
    WHERE extern.QuestionID = intern.QuestionID
    AND intern.IsCorrect=1
    ORDER BY Choice
    FOR XML PATH('')
  ) pre_trimmed (CorrectChoice)
  GROUP BY QuestionID, CorrectChoice
)
, Answer AS
(
  SELECT QuestionID, UserName, LEFT(CorrectChoice , LEN(CorrectChoice )-1) AS Choice
  FROM SurveyAnswers AS extern
  CROSS APPLY
  (
    SELECT Choice + ', '
    FROM SurveyAnswers intern
    INNER JOIN SurveyChoices SC ON SC.QuestionId = intern.QuestionID
                               AND SC.ChoiceID = intern.ChoiceID
    WHERE extern.QuestionID = intern.QuestionID
    ORDER BY Choice
    FOR XML PATH('')
  ) pre_trimmed (CorrectChoice)
  GROUP BY QuestionID, UserName, CorrectChoice
) 

SELECT SQ.Question, CC.Choice, A.Choice Answer, 
       CASE WHEN CC.Choice = A.Choice THEN 'Correct' ELSE 'Wrong' END AS Status
FROM SurveyQuestions SQ
LEFT JOIN CorrectChoices CC ON CC.QuestionID = SQ.QuestionID
LEFT JOIN Answer A ON A.QuestionID =SQ.QuestionID
WHERE SQ.SurveyID = @SurveyId AND UserName=@UserName">
<SelectParameters>
 <asp:SessionParameter Name="UserName" SessionField="UserName" />
 <asp:SessionParameter Name="SurveyId" SessionField="SurveyId" />
</SelectParameters>
</asp:SqlDataSource>
4

1 回答 1

0

我已经解决了这个问题。

只需添加 intern.UserName = @Username 即可解决内部查询的问题。

SELECT Choice + ', ' FROM SurveyAnswers intern INNER JOIN SurveyChoices SC ON SC.QuestionId = intern.QuestionID AND SC.ChoiceID = intertern.ChoiceID和 intern.UserName = @UserName WHERE extern.QuestionID = intern.QuestionID

于 2013-09-16T15:19:31.237 回答