Question

我有一个如下的SQL查询

SELECT Count(L.id) As Count, U.name As Name, L.action As Action
FROM tabel1 L LEFT JOIN tabel2 U ON L.uId = U.id
WHERE L.date >= '1/1/2018 12:00:00 AM' AND L.date <= '11/2/2019 12:00:00 AM' AND L.nId IN (18161, 18156, 4924, 18129, 18073, 18070, 18140, 2588, 18141, 18139, 18144, 18143, 18142, 6741, 16796, 3333, 6364, 6367, 6370, 6373, 17991, 7812, 11148, 7799, 11074) 
group by U.name, L.action
ORDER BY U.name, L.action

其结果如下

Count   Name        Action
--------------------------     
24      User 1      Create
31      User 1      Edit
252     User 2      Create
48      User 2      Edit
257     User 3      Create
29      User 3      Edit
3       User 4      Create
216     User 5      Create
287     User 5      Edit
256     User 6      Edit
9       User 6      Create
1       User 6      Delete
101     User 7      Edit

我需要得到如下结果

Name      Create      Edit      Delete
--------------------------------------
User 1    24          31        0
User 2    252         48        0
User 3    257         29        0
User 4    3           0         0
User 5    216         287       0
User 6    256         9         1
User 7    0           101       0

我该如何实现？

Answer 1

只需使用条件聚合：

SELECT U.name As Name, 
       SUM(CASE WHEN l.Action = 'Create' THEN 1 ELSE 0 END) as num_created,
       SUM(CASE WHEN l.Action = 'Edit' THEN 1 ELSE 0 END) as num_edited,
       SUM(CASE WHEN l.Action = 'Delete' THEN 1 ELSE 0 END) as num_deleted
FROM tabel1 L LEFT JOIN
     tabel2 U
     ON L.uId = U.id
WHERE L.date >= '2018-01-01' AND
      L.date <= '2019-02-11' AND
      L.nId IN (18161, 18156, 4924, 18129, 18073, 18070, 18140, 2588, 18141, 18139, 18144, 18143, 18142, 6741, 16796, 3333, 6364, 6367, 6370, 6373, 17991, 7812, 11148, 7799, 11074) 
GROUP BY U.name
ORDER BY U.name;

请注意，我还简化了日期比较并介绍了ISO标准格式。但是，我不确定以后的日期是2019-02-11还是2019-11-02。

如何使用子查询或数据透视来反向查询选择

1 个答案: