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我需要您的帮助来创建查询。我是软件职业的新手，我正在尝试为 Windows Mobile 应用程序开发 KBC 类型测验游戏。为此，我想尝试从数据库中获取随机记录，如果一条记录出现一次，则该记录将不再出现。我使用此查询"SELECT TOP 1 * FROM Quiz ORDER BY NEWID()"，但记录重复。

我的表结构如下所示。

ColumnName  DataType

Id          int
Que         varchar(150)
Ans1        varchar(100)
Ans2        varchar(100)
Ans3        varchar(100)
TrueAns     varchar(100)

我也尝试创建一个存储过程，如下所示

DECLARE @counter int, @randno int, @uBound int, @lBound int


        SELECT @uBound = Max(Id) FROM Quiz
        SELECT @lBound = Min(Id) FROM Quiz

        SELECT @randno = Round(((@uBound - @lBound) * Rand() + @lBound), 0)
        SET @Counter = 0

        WHILE @counter = 0
        BEGIN
            IF EXISTS(SELECT Id FROM Quiz WHERE Id = @randno)
            BEGIN
                SET NOCOUNT OFF
                SELECT * FROM Quiz WHERE Id = @randno
                SET @counter = 1
            END
            ELSE
            BEGIN
                SELECT @randno = Round(((@uBound - @lBound -1 ) * Rand() + @lBound), 0)
            END
        END

但我无法获得成功。我的表包含此字段 Que、Ans1、Ans2、Ans3、TrueAns。请帮我解决这个问题。我还想创建一个返回所有记录的网络服务

4

1 回答 1

1

只需在 Quiz isread 中添加一列作为 bool 并将其设置为默认 false

然后使您的存储过程像它一样可以帮助您使用Web服务方法

DECLARE @counter int, @randno int, @uBound int, @lBound int


        SELECT @uBound = Max(Id) FROM Quiz
        SELECT @lBound = Min(Id) FROM Quiz

        SELECT @randno = Round(((@uBound - @lBound) * Rand() + @lBound), 0)
        SET @Counter = 0

        WHILE @counter = 0
        BEGIN


If(( select count(*) from Quiz where isread= true)=( select count(*) from Quiz ))
BEGIN
 update Quiz SET isread=false  
End

            IF EXISTS(SELECT Id FROM Quiz WHERE Id = @randno and isread=false )
            BEGIN
            update Quiz SET isread=true  WHERE Id = @randno                   
            SET NOCOUNT OFF
                SELECT * FROM Quiz WHERE Id = @randno

                SET @counter = 1
            END
            ELSE
            BEGIN
                SELECT @randno = Round(((@uBound - @lBound -1 ) * Rand() + @lBound), 0)
            END
        END
于 2013-04-04T07:13:44.243 回答