我很难找到有效的解决方案来解决我的问题。我认为这是一个相当普遍的问题，这就是我在这里提出问题的原因。
这就是问题所在。
假设我有多个选择，例如以下几个：
选择创建
| date | num_created |
|---|---|
| 01-01-2021 | 10 |
| 01-02-2021 | 2 |
| 01-04-2021 | 13 |
SELECT Update
| date | num_update |
|---|---|
| 01-01-2021 | 14 |
| 01-02-2021 | 2 |
| 01-03-2021 | 9 |
SELECT Delete
| date | num_delete |
|---|---|
| 01-02-2021 | 2 |
| 01-05-2021 | 40 |
我想要这个最终输出
Final output
| date | num_created | num_update | num_deleted |
|---|---|---|---|
| 01-01-2021 | 10 | 14 | 0 |
| 01-02-2021 | 2 | 2 | 2 |
| 01-03-2021 | 0 | 9 | 0 |
| 01-04-2021 | 13 | 0 | 0 |
| 01-05-2021 | 0 | 0 | 40 |
*我不能假设任何表格都有所有日期或匹配日期
【问题讨论】：
您似乎想要按天分组计数：
select
tbls.date,
sum(tbls.num_created) as num_created,
sum(tbls.num_updated) as num_updated,
sum(tbls.num_deleted) as num_deleted
from (
select date, num_created, 0 as num_updated, 0 as num_deleted from tbl_create
union all
select date, 0 as num_created, num_updated, 0 as num_deleted from tbl_update
union all
select date, 0 as num_created, 0 as num_updated, num_deleted from tbl_delete) as tbls
group by tbls.date
请不要在不需要时滥用“with 语句”（如其他答案所示）。
编辑 - 额外的简单查询：
if object_id('tempdb..#tmp_dml_statement') is not null drop table #tmp_dml_statement
select date, num_created, 0 as num_updated, 0 as num_deleted
into #tmp_dml_statement
from tbl_create
union all
select date, 0 as num_created, num_updated, 0 as num_deleted
from tbl_update
union all
select date, 0 as num_created, 0 as num_updated, num_deleted
from tbl_delete
【讨论】：
您可以在表和case语句之间执行完全外连接以选择如下日期（假设3个表/查询为a，b，c）：
With temp as
(
select case when a.date is not null then a.date else b.date end as date,num_created,num_update from a full join b on (a.date=b.date)
),
temp1 as
(
select case when temp.date is not null then temp.date else c.date end as date,num_created,num_update,num_delete from temp full join c on (temp.date=c.date)
)
select * from temp1;
【讨论】：
假设表格是：
tbl_create(dt, num_created)
tbl_update(dt, num_updated)
tbl_delete(dt, num_deleted)
从包含 3 个源表中所有日期的（临时）表开始：all_dates
然后（左外）加入 3 个源表中的计数器。
注意：'union' 删除重复项，无需担心（与 'union all' 相比）
注意：coalesce(a,b) 返回第一个非空参数，所以你有零而不是 'null's
with all_dates(dt) as (
select dt from tbl_create
union
select dt from tbl_update
union
select dt from tbl_delete
)
select all_dates.dt
, coalesce(tbl_create.num_created,0) --
, coalesce(tbl_update.num_updated,0)
, coalesce(tbl_delete.num_deleted,0)
from all_dates
left outer join tbl_create on all_dates.dt = tbl_create.dt
left outer join tbl_update on all_dates.dt = tbl_update.dt
left outer join tbl_delete on all_dates.dt = tbl_delete.dt
【讨论】：