加入其他 table 并在一个列中使用 group by 计算相同的 id

Question

我有四个这样的table

Laporan_Kondisi

+----+------------+----------------+
| id | id_kondisi | id_sub_kondisi |
+----+------------+----------------+
| 37 | 01         | 0102           |
| 38 | 03         | 0302           |
| 40 | 01         | 0101           |
| 41 | 01         | 0102           |
| 42 | 01         | 0101           |
| 43 | 03         | 0301           |
| 44 | 03         | 0303           |
| 45 | 02         | 0202           |
| 46 | 01         | 0102           |
| 47 | 03         | 0301           |
| 48 | 01         | 0101           |
| 49 | 02         | 0203           |
| 50 | 03         | 0302           |
| 51 | 02         | 0202           |
| 52 | 02         | 0201           |
| 53 | 02         | 0202           |
+----+------------+----------------+

Laporan_Sebab

+----+--------------------+
| id | id_laporan_kondisi |
+----+--------------------+
|  4 | 195                |
|  5 | 34                 |
|  6 | 195                |
|  7 | 37                 |
|  8 | 37                 |
| 10 | 38                 |
| 11 | 36                 |
| 12 | 40                 |
+----+--------------------+

Laporan_Rekomendasi

+----+------------------+
| id | id_laporan_sebab |
+----+------------------+
|  4 | 4                |
|  5 | 4                |
|  6 | 5                |
|  7 | 7                |
|  8 | 7                |
|  9 | 8                |
| 10 | 8                |
| 12 | 10               |
| 13 | 10               |
| 14 | 12               |
| 15 | 12               |
+----+------------------+

Laporan Tindak Lanjut

+----+------------------------+-------------------+
| id | id_laporan_rekomendasi | tgl_tindak_lanjut |
+----+------------------------+-------------------+
|  9 |                      4 | 2017-09-13        |
| 10 |                      5 | 2017-09-13        |
| 11 |                      5 | 2017-09-13        |
| 12 |                      6 | 2017-09-13        |
| 13 |                      7 | 2017-09-13        |
| 14 |                      8 | 2017-09-13        |
| 15 |                      7 | 2017-09-13        |
| 18 |                     12 | 2017-09-14        |
| 19 |                     13 | 2017-09-14        |
| 24 |                     15 | 2017-09-14        |
| 28 |                     15 | 2017-09-14        |
| 29 |                     14 | 2017-09-14        |
| 30 |                     14 | 2017-09-14        |
+----+------------------------+-------------------+

我用LEFT OUTER JOIN得到了这样的结果

查询

SELECT 
    a.id as a,  
    a.id_kondisi as id_k,
    a.id_sub_kondisi as id_s_k,
    d.tgl_tindak_lanjut as tgl_tl 
FROM `laporan_kondisi` a 
LEFT OUTER JOIN 
    laporan_sebab b 
ON 
    a.id = b.id_laporan_kondisi 
LEFT OUTER JOIN 
    laporan_rekomendasi c 
ON 
    b.id = c.id_laporan_sebab 
LEFT OUTER join 
    laporan_tindak_lanjut d 
ON 
    c.id = d.id_laporan_rekomendasi ORDER by a.id_kondisi ASC,a.id_sub_kondisi ASC

结果

a   id_k    id_s_k  tgl_tl
48    01    0101    2017-09-14
40    01    0101    2017-09-14
40    01    0101    2017-09-13
40    01    0101    2017-09-14
42    01    0101    2017-09-13
40    01    0101    2017-09-14
37    01    0102    2017-09-14
37    01    0102    2017-09-14
46    01    0102    2017-09-14
37    01    0102    2017-09-14
41    01    0102    2017-09-13
37    01    0102    2017-09-13
52    02    0201    2017-09-14
45    02    0202    2017-09-14
51    02    0202    2017-09-14
53    02    0202    2017-09-13
49    02    0203    2017-09-14
47    03    0301    2017-09-14
43    03    0301    2017-09-13
38    03    0302    2017-09-13
38    03    0302    2017-09-13
50    03    0302    2017-09-13
44    03    0303    2017-09-13

您可以在 SQLFiddle 上看到结果 table。我可以像我给的 fiddle 那样加入四个 table。但同时我想知道如何像下面的table一样加入和计数：

id_k    count_09_14   count_09_13
01         8                4
0101       4                2
0102       4                2
02         4                1
0201       1                0
0202       2                1  
0203       1                0
03         1                5
0301       1                1
0302       0                3
0303       0                1

所以我想要实现的是，id_s_k 将加入 id_k，然后在 count_09_14 中将计算基于 tgl_tl 循环的 ID whic 将计算循环 id_k 和 id_s_k，其中 tgl_tl 是 2017-09-14。与 count_09_13 相同。它将从 id_k 和 id_s_k 计算相同的 ID，其中 tgl_tl 是 2017-09-13。我该怎么做？

感谢

Answer 1

终于，我找到了解决方案，如果有人有其他解决方案（同时使用查询或在应用程序代码中），我将不胜感激

SQLFiddle

SELECT  
   id,
    SUM(CASE WHEN tgl_tindak_lanjut="2017-09-14" THEN 1 ELSE 0 end ) as count_all_09,
   SUM( CASE WHEN tgl_tindak_lanjut="2017-09-13" THEN 1 ELSE 0 END) as count_09_13

FROM (
  SELECT 
    a.id_kondisi as id,
    d.tgl_tindak_lanjut as tgl_tindak_lanjut
  FROM 
    laporan_kondisi a
  LEFT OUTER JOIN 
    laporan_sebab b 
ON 
    a.id = b.id_laporan_kondisi 
LEFT OUTER JOIN 
    laporan_rekomendasi c 
ON 
    b.id = c.id_laporan_sebab 
LEFT OUTER join 
    laporan_tindak_lanjut d 
ON 
    c.id = d.id_laporan_rekomendasi
  UNION ALL SELECT 
    a2.id_sub_kondisi,
    d2.tgl_tindak_lanjut as tgl_tindak_lanjut
  FROM laporan_kondisi a2
  LEFT OUTER JOIN 
    laporan_sebab b2
ON 
    a2.id = b2.id_laporan_kondisi 
LEFT OUTER JOIN 
    laporan_rekomendasi c2 
ON 
    b2.id = c2.id_laporan_sebab 
LEFT OUTER join 
    laporan_tindak_lanjut d2 
ON 
    c2.id = d2.id_laporan_rekomendasi
  ) merged_table

GROUP BY id

加入其他 table 并在一个列中使用 group by 计算相同的 id

Join other table and count the same id with group by in one single column

php

mysql

union

sum

outer-join