select * from ( select uid ,min(date(in_time)) dt from tb_user_log group by uid) as t1 left join (select uid , date(in_time) dt from tb_user_log union select uid , date(out_time) from tb_user_log) as t2 语法检查
时间: 2024-04-08 08:34:53 浏览: 12
这个SQL语句会检查语法是否正确。以下是经过修正的查询语句:
```sql
SELECT *
FROM (
SELECT uid, MIN(DATE(in_time)) AS dt
FROM tb_user_log
GROUP BY uid
) AS t1
LEFT JOIN (
SELECT uid, DATE(in_time) AS dt
FROM tb_user_log
UNION
SELECT uid, DATE(out_time)
FROM tb_user_log
) AS t2;
```
请注意,这个查询会从 `tb_user_log` 表中选择每个用户的最早登录日期,并将其与所有登录和登出日期进行联接。
相关问题
with t1 AS (select uid ,min(date(in_time)) dt from tb_user_log3 group by uid ) t2 AS (select uid,date(in_time) dt from tb_user_log3) select t1.dt,round(count(t2.uid)/count(t1.uid),2) uv_rate from t1 left join t2 on t1.uid=t2.uid and t1.dt=date_sub(t2.dt,INTERVAL 1 day) where date_format(t1.dt,'%Y-%m') = '2021-11' group by t1.dt order by t1.dt 有什么语法错误
你的查询语句存在一些语法错误。在使用WITH子句创建临时表时,应该使用逗号将不同的临时表分隔开。此外,临时表的定义应该放在查询语句的开始部分。
以下是修正后的查询语句:
```sql
WITH t1 AS (
SELECT uid, MIN(DATE(in_time)) AS dt
FROM tb_user_log3
GROUP BY uid
),
t2 AS (
SELECT uid, DATE(in_time) AS dt
FROM tb_user_log3
)
SELECT t1.dt, ROUND(COUNT(t2.uid)/COUNT(t1.uid), 2) AS uv_rate
FROM t1
LEFT JOIN t2 ON t1.uid = t2.uid AND t1.dt = DATE_SUB(t2.dt, INTERVAL 1 DAY)
WHERE DATE_FORMAT(t1.dt, '%Y-%m') = '2021-11'
GROUP BY t1.dt
ORDER BY t1.dt;
```
在修正后的查询语句中,使用了逗号将t1和t2两个临时表分隔开,并将临时表的定义放在查询语句的开始部分。
请注意,修正后的查询语句假设数据库中存在名为`tb_user_log3`的表,并且表中包含`uid`和`in_time`列。如果你的数据库结构不同,请根据实际情况进行调整。
select a.dt ,sum(case when b.time1 = date_add(a.dt,1) then 1 else 0 end) as `uv_left_rate` from (select uid,min(date(in_time)) dt from tb_user_log group by uid having dt like '2021-11%') a left join (select uid,date(in_time) time1 from tb_user_log group by uid,time1 ) b on a.uid = b.uid group by dt order by dt存在什么问题
你这个 SQL 语句中的问题是在第1个子查询中,select 语句中的 dt 列不能直接使用 like 来进行过滤,因为 dt 列的数据类型是日期,不是字符串。正确的做法是使用 date_format 函数来将日期类型转换为字符串类型,然后再使用 like 进行比较,例如:
```
select uid, min(date(in_time)) dt
from tb_user_log
where date_format(in_time, '%Y-%m') = '2021-11'
group by uid
```
另外,你的第2个子查询中也存在问题,因为在 group by 子句中,uid 和 time1 两列都出现,但是 time1 列并没有做任何聚合函数的计算,这样会导致结果集中出现重复的 uid 和 time1 组合。正确的做法是只按照 uid 分组,例如:
```
select uid, date(in_time) time1
from tb_user_log
group by uid, time1
```
综上所述,修正后的 SQL 语句应该是这样的:
```
select a.dt, sum(case when b.time1 = date_add(a.dt, 1) then 1 else 0 end) as `uv_left_rate`
from (
select uid, min(date(in_time)) dt
from tb_user_log
where date_format(in_time, '%Y-%m') = '2021-11'
group by uid
) a
left join (
select uid, date(in_time) time1
from tb_user_log
group by uid, date(in_time)
) b on a.uid = b.uid
group by a.dt
order by a.dt
```