Question

我有这样的查询（我在两个查询中都有join和union的复杂查询）

SELECT 
            id, 
            COUNT(id) ct_id, 
            SUM(kl) sum_kl, 
            SUM(CASE WHEN tgl_kondisi="2017-09-13" THEN 1 ELSE 0 END) as tot, 
            SUM(CASE WHEN tgl_kondisi <= "2017-09-14" THEN kl ELSE 0 end ) as sum_kl2 
        FROM (
            SELECT 
                id_kondisi as id, 
                tgl_kondisi,
                nilai_potensi_kerugian AS kl 
            FROM laporan_kondisi 
            UNION  
            SELECT 
                id_sub_kondisi, 
                tgl_kondisi, 
                nilai_potensi_kerugian 
            FROM laporan_kondisi 
        ) merged_table  GROUP BY id ORDER BY id;

结果

+------+-------+----------+------+----------+
| id   | ct_id | sum_kl   | tot  | sum_kl2  |
+------+-------+----------+------+----------+
| 01   |     3 | 20000000 |    3 | 20000000 |
| 0101 |     2 |  9000000 |    2 |  9000000 |
| 0102 |     2 | 11000000 |    2 | 11000000 |
| 02   |     1 |        0 |    0 |        0 |
| 0201 |     1 |        0 |    0 |        0 |
| 0202 |     1 |        0 |    0 |        0 |
| 0203 |     1 |        0 |    0 |        0 |
| 03   |     2 |  4000000 |    2 |  4000000 |
| 0301 |     1 |        0 |    1 |        0 |
| 0302 |     2 |  4000000 |    2 |  4000000 |
| 0303 |     1 |        0 |    1 |        0 |
+------+-------+----------+------+----------+

另一个问题

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  
        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

结果

+------+--------------+-------------+
| id   | count_all_09 | count_09_13 |
+------+--------------+-------------+
| 01   |            0 |           1 |
| 0101 |            0 |           1 |
| 0102 |            0 |           0 |
| 02   |            0 |           0 |
| 0201 |            0 |           0 |
| 0202 |            0 |           0 |
| 0203 |            0 |           0 |
| 03   |            0 |           0 |
| 0301 |            0 |           0 |
| 0302 |            0 |           0 |
| 0303 |            0 |           0 |
+------+--------------+-------------+

这样可以将表格组合起来吗？

+------+-------+----------+------+----------+--------------+-------------+
| id   | ct_id | sum_kl   | tot  | sum_kl2  | count_all_09 | count_09_13 |
+------+-------+----------+------+----------+--------------+-------------+
| 01   |     3 | 20000000 |    3 | 20000000 |     0        |       1     |
| 0101 |     2 |  9000000 |    2 |  9000000 |     0        |       1     |
| 0102 |     2 | 11000000 |    2 | 11000000 |     0        |       0     |
| 02   |     1 |        0 |    0 |        0 |     0        |       0     |
| 0201 |     1 |        0 |    0 |        0 |     0        |       0     |
| 0202 |     1 |        0 |    0 |        0 |     0        |       0     |
| 0203 |     1 |        0 |    0 |        0 |     0        |       0     |
| 03   |     2 |  4000000 |    2 |  4000000 |     0        |       0     |
| 0301 |     1 |        0 |    1 |        0 |     0        |       0     |
| 0302 |     2 |  4000000 |    2 |  4000000 |     0        |       0     |
| 0303 |     1 |        0 |    1 |        0 |     0        |       0     |
+------+-------+----------+------+----------+--------------+-------------+

我尝试了JOIN和许多其他但只提供一个结果。这可能就像这样做吗？或者这个查询只是给我们没有结果？

Answer 1

很难说通过添加所有第二个查询逻辑，结果是否会受到影响，因为您没有提供有关此查询执行的操作或实际数据的任何信息。您可以做什么，哪些应该始终有效（如果您计划重新使用此查询，则不推荐，如果性能是一个问题，也不建议这样做），只需加入整个查询即可。

看起来像这样：

SELECT t.*,s.count_all_09,s.count_09_13
FROM (FIRST QUERY HERE) t
JOIN (SECOND QUERY HERE) s
 ON(t.id = s.id)

将两个结果查询合并为一个

1 个答案: