Php MSSQL:必须声明标量变量@rownum

Php MSSQL:必须声明标量变量@rownum,php,sql-server,variables,scalar,Php,Sql Server,Variables,Scalar,我的一个问题是一个游戏的排名 下面是查询,SQL Server Management Studio显示的错误是:“必须声明标量变量@rownum” 怎么了 非常感谢 $sql1_1 = "SET @rownum := 0"; $sql2_2 = "SELECT * FROM ( SELECT @rownum := @rownum + 1 AS rank, totalpoints, useridFB, game2point

我的一个问题是一个游戏的排名

下面是查询,SQL Server Management Studio显示的错误是:“必须声明标量变量@rownum”

怎么了

非常感谢

 $sql1_1     = "SET @rownum := 0";



$sql2_2       =   "SELECT * FROM (
                      SELECT @rownum := @rownum + 1 AS rank, totalpoints, useridFB, game2points
                      FROM theuser ORDER BY game2points DESC
                      ) as result WHERE useridFB=1234";



        mssql_query($sql1_1); 
        $result = mssql_query($sql2_2);

        $row = mssql_fetch_array($result);
        $therank = $row['rank'];

您正在SQL Server中使用MySql语法

使用
row\u number()
函数再现当前逻辑

SELECT *
FROM   (SELECT row_number() OVER (ORDER BY game2points DESC) AS [rank],
               totalpoints,
               useridFB,
               game2points
        FROM   theuser) AS result
WHERE  useridFB = 1234  

或者,您可能希望根据希望如何处理关系来调查
排名

您在SQL Server中使用的是MySql语法

使用
row\u number()
函数再现当前逻辑

SELECT *
FROM   (SELECT row_number() OVER (ORDER BY game2points DESC) AS [rank],
               totalpoints,
               useridFB,
               game2points
        FROM   theuser) AS result
WHERE  useridFB = 1234  

或者,您可能希望根据您希望如何处理领带来调查
排名

@user-No。您根本不需要设置
或变量。@user-No。您根本不需要设置
或变量。