我有这个问题:
$query = "
SET @points := -1;
SET @num := 0;
SELECT `id`,`rank`,
@num := if(@points = `rank`, @num, @num + 1) as `point_rank`
FROM `said`
ORDER BY `rank` *1 desc, `id` asc";
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 ‘SET @num := 0;
如果我在PHPmyadmin SQL查询面板中复制并粘贴该代码,它可以正常工作,但是从PHP代码行开始它不起作用,看起来设置Vars时会出现问题.
解决方法:
您是否尝试使用CROSS JOIN而不是在单独的SET中设置变量:
$query = "
SELECT `id`,
`rank`,
@num := if(@points = `rank`, @num, @num + 1) as `point_rank`
FROM `said`
CROSS JOIN (SELECT @points:=-1, @num:=0) c
ORDER BY `rank` *1 desc, `id` asc";
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