语句如下
一.
SELECT top 2 user_Integral_tbl.user_Integral_customers_id,
sum( user_Integral_tbl.user_Integral_Numerical) as 总成绩
FROM user_Integral_tbl
where not exists (SELECT top 2 user_Integral_tbl.user_Integral_customers_id FROM user_Integral_tbl GROUP BY user_Integral_tbl.user_Integral_customers_id ORDER BY sum( user_Integral_tbl.user_Integral_Numerical) DESC)
GROUP BY user_Integral_tbl.user_Integral_customers_id
ORDER BY 总成绩 DESC
二.
SELECT top 2 user_Integral_tbl.user_Integral_customers_id,
sum( user_Integral_tbl.user_Integral_Numerical) as 总成绩
FROM user_Integral_tbl
where user_Integral_tbl.user_Integral_customers_id not in (SELECT top 2 user_Integral_tbl.user_Integral_customers_id FROM user_Integral_tbl GROUP BY user_Integral_tbl.user_Integral_customers_id ORDER BY sum( user_Integral_tbl.user_Integral_Numerical) DESC)
GROUP BY user_Integral_tbl.user_Integral_customers_id
ORDER BY 总成绩 DESC
------解决方案--------------------
因为一里面的not exists(子查询),子查询总是有结果返回,如果子查询没有结果返回,同样的查询也没有结果返回
------解决方案--------------------
修改
一.
SELECT top 2 user_Integral_tbl.user_Integral_customers_id,
sum( user_Integral_tbl.user_Integral_Numerical) as 总成绩
FROM user_Integral_tbl
GROUP BY user_Integral_tbl.user_Integral_customers_id
ORDER BY 总成绩 DESC