Mysql 使用两个条件获取计数的嵌套查询

我有这三张桌子

tbl_1-
ip  |isp    |infection
----------------------
1   |aaaa   |malware
2   |bbbb   |malware
3   |cccc   |ddos
3   |cccc   |trojan
4   |dddd   |ddos

tbl_2-
ip  |isp    |infection
----------------------
1   |aaaa   |malware
3   |cccc   |ddos
4   |dddd   |trojan
5   |eeee   |trojan
6   |ffff   |other

tbl_3-
ip  |isp    |infection
----------------------
1   |aaaa   |ddos
6   |ffff   |
2   |bbbb   |other

我需要得到如下结果

result-
ip  |isp    |infection  |ipCount    |ispCount   |infectionCount
--------------------------------------------------------------
1   |aaaa   |malware    |3          |3          |2
1   |aaaa   |ddos       |3          |3          |1
2   |bbbb   |other      |2          |2          |1
2   |bbbb   |malware    |2          |2          |1
3   |cccc   |ddos       |3          |3          |2
3   |cccc   |trojan     |3          |3          |1
4   |dddd   |ddos       |2          |2          |1
4   |dddd   |trojan     |2          |2          |1
5   |eeee   |trojan     |1          |1          |1
6   |ffff   |other      |2          |2          |1
6   |ffff   |           |2          |2          |1

ipCount, ispCount -> count of matching ip and isp
    eg-there are 3 records with ip = 1 and isp = aaaa

infectionCount -> count of matching infections per ip and isp
    eg-there are 2 infections that says malware where ip = 1 and isp = aaaa

我想我需要一个嵌套的查询，但我不知道如何在两个条件下计数；你能帮忙吗

编辑：我尝试过的代码

SELECT ip, isp, infection, count(ip), count(isp), count(infection)
FROM (

SELECT ip, isp, infection
FROM tbl_1
UNION ALL
SELECT ip, isp, infectionType
FROM tbl_2
UNION ALL
SELECT ip, isp, infection
FROM tbl_3
)x

GROUP BY ip, isp, infection

---

但是它没有给出我想要的结果，因为我不知道如何在一个查询中进行两种类型的计数

您可以将所有表合并在一起，然后对列进行求和，并按特定列进行分组

SELECT ip, isp, infection, COUNT(ip) AS ipcount, COUNT(isp) AS ispcount, COUNT(infection) AS infectioncount
FROM
  (
    SELECT ip, isp, infection
    FROM table_1
    UNION ALL
    SELECT ip, isp, infection
    FROM table_2
    UNION ALL
    SELECT ip, isp, infection
    FROM table_3
  )
GROUP BY ip, isp, infection
ORDER BY ip, isp, infection;

---

您需要对列

infection

和（

ip

&

ipc

）进行不同的分组，然后使用如下子查询将它们连接起来：

SELECT t1.ip, t1.isp, t2.infection, t1.ipc, t1. ispc, t2.incount
FROM
    (SELECT ip, isp, infection, COUNT(ip) as ipc, COUNT(isp) as ispc
    FROM (
       SELECT ip, isp, infection
       FROM tbl1
       UNION ALL
       SELECT ip, isp, infection
       FROM tbl2
       UNION ALL
       SELECT ip, isp, infection
       FROM tbl3
       )x
     GROUP BY ip, isp) t1
JOIN
    (SELECT ip, isp, infection, COUNT(infection) as incount
     FROM (
       SELECT ip, isp, infection
       FROM tbl1
       UNION ALL
       SELECT ip, isp, infection
       FROM tbl2
       UNION ALL
       SELECT ip, isp, infection
       FROM tbl3
       )x
    GROUP BY ip, isp,  infection)t2
ON t1.ip = t2.ip
ORDER BY ip, isp, infection Desc

注意：我认为您期望的输出是错误的，因为：

在

表3

中，对于

ip=6

没有

感染

，但它在您的输出中

感染

其他

在您的输出中丢失（而是有

恶意软件

）

---

所以我尝试了联合，我会发布代码。

ipCount

与

iscount

什么时候不同？除非isp名称是different@hims056我发布了我尝试过的代码，它足以使用GROUP BY对输出进行排序。不..这与我之前尝试过的解决方案相同..但事实并非如此：（感谢回复：）嘿嘿…谢谢：）是的..我做了编辑..谢谢你指出：）