Php MSSQL:必须声明标量变量@rownum
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
$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。