level 9
where 你说的条件 group user
2018年04月05日 14点04分
level 7
5
直接这样加条件是不对的Where tm-id＝1 and daan＝a and tm-id＝2 and daan＝c 需求是查询同时答了1.2两题
2018年04月06日 08点04分
level 11
SELECT
COUNT(1)
FROM
user_detail d
INNER JOIN user_detail d2 ON d.id = d2.id
WHERE
d.tm_id = 1
AND d.da_an = 'd'
or d2.tm_id=2
and d2.da_an='a'
2018年04月05日 14点04分
6
COUNT(1)
FROM
user_detail d
INNER JOIN user_detail d2 ON d.id = d2.id
WHERE
d.tm_id = 1
AND d.da_an = 'd'
or d2.tm_id=2
and d2.da_an='a'
![[吐舌]](/static/emoticons/u5410u820c.png)
level 5
select count(1) from user_detail where concat(tm_id,da_an) in ('1b','2a')
2018年04月06日 03点04分
11
level 15
select user from userdetail a where a.tmid＝1 and a.daan＝b or a.tmid＝2 and a.daan＝a group by a.user having count (1)＝2
或者
select a.user form userdetail a，user detail b where a.user＝b.user and a.tmid＝1 and a.daan＝b and b.tmid＝2 and b.daan＝a
2018年04月06日 15点04分
12
或者
select a.user form userdetail a，user detail b where a.user＝b.user and a.tmid＝1 and a.daan＝b and b.tmid＝2 and b.daan＝a
![[冷]](/static/emoticons/u51b7.png)
level 11
![[疑问]](/static/emoticons/u7591u95ee.png)
?![[阴险]](/static/emoticons/u9634u9669.png)
我怎么感觉说不清了
level 11
select count(1) from (select
count(1)
FROM
user_d a
WHERE
a.tm_id=1
AND a.da_an='d'
OR
a.tm_id=2
AND a.da_an='a'
GROUP BY
a.`user`
HAVING
count(a.user)=2)
user_d
改成这样 没问题
select count(1) from (select
count(1)
FROM
user_d a
WHERE
a.tm_id=1
AND a.da_an='d'
OR
a.tm_id=2
AND a.da_an='a'
GROUP BY
a.`user`
HAVING
count(a.user)=2)
user_d
@abcwt112
2018年04月07日 08点04分
15
count(1)
FROM
user_d a
WHERE
a.tm_id=1
AND a.da_an='d'
OR
a.tm_id=2
AND a.da_an='a'
GROUP BY
a.`user`
HAVING
count(a.user)=2)
user_d
改成这样 没问题
select count(1) from (select
count(1)
FROM
user_d a
WHERE
a.tm_id=1
AND a.da_an='d'
OR
a.tm_id=2
AND a.da_an='a'
GROUP BY
a.`user`
HAVING
count(a.user)=2)
user_d
@abcwt112