Question

我有一张表，可以根据值存储每天用户是否工作或休假。示例表，Value = 1 -> WorkDay, Value = 2 -> Vacation：

User    | Day        | Value
--------|------------|-------
user-1  | 2021-01-01 | 1
user-1  | 2021-01-02 | 1
user-1  | 2021-01-03 | 1
user-1  | 2021-01-04 | 1
user-1  | 2021-01-05 | 2
user-1  | 2021-01-06 | 2
...

我想把这个表转换成这个（使用上面的简单例子）：

User    | Year | Month | WorkDay | Vacation
--------|------|-------|---------|---------
user-1  | 2021 | 01    | 4       | 2
...

我尝试使用 group by、子查询和 case，但整个过程变得一团糟。

SELECT
    YEAR(Day),
    MONTH(DAY),
    User,

    ...

From Table1
Group By YEAR(Day), MONTH(DAY), User

Answer 1

只需使用条件聚合：

gcloud pubsub subscriptions create supervisor-subscription \
    --topic supervisor-cron \
    --push-endpoint=$(gcloud run services describe supervisor --format 'value(status.url)' --platform managed --project ${PROJECT_ID}) \
    --push-auth-service-account=cloud-run-pubsub-invoker@${PROJECT_ID}.iam.gserviceaccount.com

Answer 2

您可以使用如下条件聚合：

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DB-小提琴：

架构和插入语句：

select User,
       year(day), 
       month(day),
       sum(case when value = 1 then 1 else 0 end) as WorkDay,
       sum(case when value = 2 then 1 else 0 end) as Vacation
from table1
group by User, year(day), month(day)

查询：

 create table table1([User] varchar(50), Day date, Value int);
 insert into table1 values('user-1'  , '2021-01-01' , 1);
 insert into table1 values('user-1'  , '2021-01-02' , 1);
 insert into table1 values('user-1'  , '2021-01-03' , 1);
 insert into table1 values('user-1'  , '2021-01-04' , 1);
 insert into table1 values('user-1'  , '2021-01-05' , 2);
 insert into table1 values('user-1'  , '2021-01-06' , 2);

输出：

<头>

用户	年	月	工作日	假期
user-1	2021	1	4	2

dbhere

通过对期间进行分组来计算记录数

2 个答案: