Question

我有一个mysql查询，有时会导致缺少值。对于我的仪表板，我想填写这些值，但如果可以的话，我希望避免建立虚拟表。

查询：

SELECT COUNT(Comms_Timestamp) as call_count,DAYOFWEEK(Comms_Timestamp) as bucket 
FROM tblTest GROUP BY bucket;

产生

+------------+--------+
| call_count | bucket |
+------------+--------+
|          4 |      1 |
|          7 |      2 |
|          7 |      3 |
|          1 |      5 |
|          6 |      6 |
|          1 |      7 |
+------------+--------+

在上面的示例中，您可以看到存储桶4丢失。我考虑了连接到选择联合数组的方法，但是由于这两个字段都是聚合的，因此我不确定该如何处理。

测试数据为

+---------------------+
| Comms_Timestamp     |
+---------------------+
| 2018-12-24 06:04:05 |
| 2018-12-24 12:18:39 |
| 2018-12-21 04:24:31 |
| 2018-12-21 08:32:44 |
| 2018-12-30 01:41:06 |
| 2018-12-30 01:53:00 |
| 2018-12-30 01:53:39 |
| 2018-12-30 02:00:01 |
| 2018-12-17 15:55:03 |
| 2018-12-17 16:04:12 |
| 2018-12-17 16:05:41 |
| 2018-12-17 16:07:43 |
| 2018-12-17 16:10:25 |
| 2018-12-18 14:03:22 |
| 2018-12-18 14:03:29 |
| 2018-12-18 14:10:19 |
| 2018-12-18 14:10:29 |
| 2018-12-18 14:10:31 |
| 2018-12-18 14:10:47 |
| 2018-12-18 14:10:55 |
| 2018-12-20 08:21:07 |
| 2018-12-28 11:03:59 |
| 2018-12-28 12:06:40 |
| 2018-12-28 12:15:01 |
| 2018-12-28 14:29:24 |
| 2019-01-05 13:33:43 |
+---------------------+

Answer 1

由于您使用的是mysql并且无权访问seq_选项，因此这是另一种方法：

SELECT A.x AS bucket, IF(ISNULL(COUNT(t2.Comms_Timestamp)), 0, COUNT(t2.Comms_Timestamp)) AS call_count FROM
(select 1 x union select 2 union select 3 union select 4 union select 5 union select 6 union select 7) AS A
LEFT JOIN tblTest AS t2 ON DAYOFWEEK(t2.Comms_Timestamp) = A.x
GROUP BY bucket
ORDER BY bucket;

这可能不是最漂亮的选择，但可以满足您的需求。

这是一个数据库字段链接：db<>fiddle

Answer 2

如果您使用的是MariaDB，则他们的Sequence Storage Engine

此表不需要创建，但是必须知道最大值。

select version();

| version()                                   |
| :------------------------------------------ |
| 10.3.11-MariaDB-1:10.3.11+maria~stretch-log |

create table bob (a int)

✓

insert into bob values (4),(2)

✓

select * from seq_1_to_5

| seq |
| --: |
|   1 |
|   2 |
|   3 |
|   4 |
|   5 |

SELECT s.seq, bob.a
FROM seq_1_to_5 s 
LEFT JOIN bob 
  ON bob.a = s.seq
ORDER BY s.seq

seq |    a
--: | ---:
  1 | null
  2 |    2
  3 | null
  4 |    4
  5 | null

db <>提琴here

Answer 3

您可以在MYSQL中使用IFNULL（）函数：

SELECT IFNULL(COUNT(C.Comms_Timestamp),0) as call_count,IFNULL(DAYOFWEEK(C.Comms_Timestamp),0) as bucket 
FROM tblCommunication as C LEFT JOIN tblCareTeam as CT on C.id_Case = CT.id_Case 
GROUP BY CT.id_Site,bucket 
HAVING CT.id_Site=8;

mysql用0填充缺失值

3 个答案: