MySQL：如何从一个表中查找日期晚于其他表的日期的记录计数

Question

There are 2 MariaDB tables:

table1

|------+----------------------+
| phone | calldate            |
|-------+---------------------+
| 123   | 2020-01-01 17:01:00 |
| 456   | 2020-01-01 17:01:00 |
| 789   | 2020-01-01 17:01:00 |
|------+---------------------+|

table2

|------+---------------------+
| phone| calldate            |
|------+---------------------+
| 123  | 2020-01-01 16:00:00 |
| 123  | 2020-01-01 17:00:00 |
| 456  | 2020-01-01 17:00:00 |
| 123  | 2020-01-01 18:00:00 |
| 456  | 2020-01-01 18:00:00 |
| 789  | 2020-01-01 18:00:00 |
|------+---------------------+

Expected result:

|-------+------+
| phone | count|
|-------+------+
| 123   | 2    |
| 456   | 1    |
|-------+------+

How to find a count of records from table2 where calldate is early than calldate from table1 group by phone?

Answer 1

How to find a count of records from table2 where calldate is later than calldate from table1 group by phone?

This sounds like a join and aggregation:

select t2.phone, count(*)
from table2 t2 join
     table1 t1
     on t2.phone = t1.phone and t2.calldate > t1.calldate
group by t2.phone;

Answer 2

With EXISTS :

select t2.phone, count(*) count
from table2 t2 
where exists (
  select 1 from table1 t1
  where t1.phone = t2.phone and t1.calldate > t2.calldate
)  
group by t2.phone

See the demo .
Results:

| phone | count    |
| ----- | -------- |
| 123   | 2        |
| 456   | 1        |

Answer 3

You can also try with a where clause if that is not working:

select count(table2.phone) from table2 t2 
full outer join table1 t1 on t2.phone=t1.phone
where t2.calldate >t1.calldate
group by t2.phone