有一个积分表，积分从高往下排，如何取出其中的一个用户的排名，积分

sql语句是

select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time

我如何在这条sql语句的基础上，查询出用户的用户名，积分。

或者如何在php中获取一下？

回复讨论(解决方案)

知道用户名么， 直接select 用户名，积分 from table where 用户名=用户名不就完事了么

??出表??。

SELECT * FORM   (    select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分    from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time  ) T WHERE username='XXX'

这种定义用户变量的写法还是很巧妙的

SELECT * FORM   (    select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分    from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time  ) T WHERE username='XXX'

这种定义用户变量的写法还是很巧妙的

SELECT * From  (    select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分    from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time  ) T WHERE username=18511337033'

这么写 ，加上where条件就会报错，没有where条件 就正常。

下面这是错误信息。

[SQL] SELECT * From  (    select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分    from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time  ) T WHERE username=18511337033'[Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''' at line 6

这是表结构

SELECT * FORM   (    select a.username,(@rowNum:=@rowNum+1) as c,a.integral as 积分    from xs_user a,(Select (@rowNum :=0 ) ) b order by integral DESC,time  ) T WHERE username='XXX'

这种定义用户变量的写法还是很巧妙的

我看错了。。是我的自己的问题。。 sorry。。   3Q  谢谢了。