set @a:=0; select id, a+b+c, @a:=@a+1 from table_name order by a+b+c desc;
加了,变成ASC排名也是错的呀,我要求得分高的排第一,怎么解决,谢谢了
已经解决了,感谢四楼的朋友提供思路set @a:=(SELECT COUNT(NAME) FROM t_name)+1; SELECT *, SUM(yuwen+suxue+english) as zongfen, @a:=@a-1 as mingci from t_name GROUP BY `name` ORDER BY zongfen DESC
有新问题了,上面的代码在实际运行中有以下提示: 我在PHP页中用 $qrs=(set @a:=(SELECT COUNT(NAME) FROM t_name)+1; SELECT *, SUM(yuwen+suxue+english) as zongfen, @a:=@a-1 as mingci from t_name GROUP BY `name` ORDER BY zongfen DESC); /*上面括号中为9楼的查询语句,在mysql中测试正常,但执行下面语句时出现错误提示*/ $rcdset=mysql_query($qrs); /*以下是错误提示*/ 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 'SELECT @a:=@a-1 as mingci, p8_form_content_63.id, p8_form_content_63.uid, p8_' at line 1这是怎么回事呀,程序应当没有错呀,请高手们指教
select id, a+b+c, @a:=@a+1
from table_name
order by a+b+c desc;
SELECT *,
SUM(yuwen+suxue+english) as zongfen,
@a:=@a-1 as mingci
from t_name
GROUP BY `name`
ORDER BY zongfen DESC
我在PHP页中用
$qrs=(set @a:=(SELECT COUNT(NAME) FROM t_name)+1;
SELECT *,
SUM(yuwen+suxue+english) as zongfen,
@a:=@a-1 as mingci
from t_name
GROUP BY `name`
ORDER BY zongfen DESC);
/*上面括号中为9楼的查询语句,在mysql中测试正常,但执行下面语句时出现错误提示*/
$rcdset=mysql_query($qrs);
/*以下是错误提示*/
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 'SELECT @a:=@a-1 as mingci, p8_form_content_63.id, p8_form_content_63.uid, p8_' at line 1这是怎么回事呀,程序应当没有错呀,请高手们指教