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MySQL两个查询的总数

[英]MySQL sum of counts of two queries

I am (very) new to MySQL. 我是MySQL的新手。 Forgive my lack of knowledge.... 原谅我缺乏知识。

I am working on a hockey stat database where I need to add all the assists and all the goals together to get "total points". 我正在曲棍球统计数据库上,我需要将所有辅助功能和所有目标加在一起以获得“总分”。 I have two queries already figured out, but I am not able to figure out how to sum the two. 我已经找到了两个查询,但是我无法弄清楚如何对两个查询求和。 Here are the queries: 以下是查询:

select player_id, count(*) 
from(select * from 1st_assists
union
select * from 2nd_assists) as tem
join players on tem.fk_player_id=players.player_id
group by fk_player_id
order by count(*) desc

select player_id, count(*)
from goals_for
join shots_for on goals_for.fk_shot_for_id=shots_for.shot_for_id
join players on shots_for.fk_player_id=players.player_id
group by player_id
order by count(*) desc;

how do I combine these two queries into one and get the total of both counts? 如何将这两个查询合并为一个并获得两个计数的总和?

Here are the results of each of the queries 这是每个查询的结果

Total Assists
player_id   count(*)
79      24
55      22
45      17
90      16
40      15
65      15
37      13
1       13
20      11
84      11
64      10
27      9
93      7
8       5
24      3
57      1

Goals
player_id   count(*)
90      38
37      28
40      19
55      13
45      11
1       8
24      8
20      8
84      8
27      6
8       5
79      4
65      4
93      1
64      1

It is untested, but can you, please, try this: 它未经测试,但是可以,请尝试以下操作:

select p.player_first_name, p.player_last_name, (count1+count2) as total_count
from
(select player_id, count(*) count1
from(select * from 1st_assists
union
select * from 2nd_assists) as tem
join players on tem.fk_player_id=players.player_id
group by fk_player_id
order by count(*) desc) q1
left join
(select player_id, count(*) count2
from goals_for
join shots_for on goals_for.fk_shot_for_id=shots_for.shot_for_id
join players on shots_for.fk_player_id=players.player_id
group by player_id) q2
ON q1.player_id=q2.player_id
left join player p ON q1.player_id=p.player_id
order by (count1+count2) desc;

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