MySQL-加入默认值

时间：2018-11-12 05:18:10

标签： mysql

我有来自MySql语句的标准数据

SELECT tipe.* 
  FROM 
     ( SELECT 'Menunggu Approve Atasan' AS tipe UNION
       SELECT 'Sudah Approved Atasan' UNION
       SELECT 'Dalam Proses Pengerjaan IT' UNION
       SELECT 'Sudah Selesai Tapi Belum Ditutup User' UNION
       SELECT 'Ditutup'
      ) AS tipe;
+---------------------------------------+
| tipe                                  |
+---------------------------------------+
| Menunggu Approve Atasan               |
| Sudah Approved Atasan                 |
| Dalam Proses Pengerjaan IT            |
| Sudah Selesai Tapi Belum Ditutup User |
| Ditutup                               |
+---------------------------------------+
5 rows in set (0.02 sec)

然后我有这样的数据

SELECT requests.* 
  FROM 
     ( SELECT DATE(request.tanggal_permintaan) as tanggal_permintaan
       , request.`status`
       , COUNT(*) as total
    FROM request
   WHERE YEAR(request.tanggal_permintaan) = '2018'
   GROUP 
    BY DATE(request.tanggal_permintaan)
       , request.status
      ) AS requests;
    +--------------------+---------------------------------------+-------+
    | tanggal_permintaan | status                                | total |
    +--------------------+---------------------------------------+-------+
    | 2018-01-02         | Menunggu Approve Atasan               |     1 |
    | 2018-01-02         | Ditutup                               |     4 |
    | 2018-01-03         | Ditutup                               |     1 |
    +--------------------+---------------------------------------+-------+
3 rows in set (0.02 sec)

您可以看到第二个查询的表单结果，它为我提供了基于日期的数据。但是，我需要在第二个查询中包括所有标准数据。 我需要这样的结果：

+--------------------+---------------------------------------+-------+
| tanggal_permintaan | status                                | total |
+--------------------+---------------------------------------+-------+
| 2018-01-02         | Menunggu Approve Atasan               |     1 |
| 2018-01-02         | Sudah Approved Atasan                 |     0 |
| 2018-01-02         | Dalam Proses Pengerjaan IT            |     0 |
| 2018-01-02         | Sudah Selesai Tapi Belum Ditutup User |     0 |
| 2018-01-02         | Ditutup                               |     4 |  // one day
| 2018-01-03         | Menunggu Approve Atasan               |     0 |
| 2018-01-03         | Sudah Approved Atasan                 |     0 |
| 2018-01-03         | Dalam Proses Pengerjaan IT            |     0 |
| 2018-01-03         | Sudah Selesai Tapi Belum Ditutup User |     0 |
| 2018-01-03         | Ditutup                               |     1 | // one day
+--------------------+---------------------------------------+-------+

我用来获得这样的查询是什么？ 到目前为止，我只是尝试了左外部联接，但是没有运气

SELECT requests.* FROM (
    SELECT
        DATE(request.tanggal_permintaan) as tanggal_permintaan,
            request.`status`,
            COUNT(*) as total
    FROM request
    WHERE YEAR(request.tanggal_permintaan) = '2018'
    GROUP BY DATE(request.tanggal_permintaan), request.status
) requests

LEFT OUTER JOIN (
    SELECT tipe.* FROM(
        SELECT 'Menunggu Approve Atasan' AS tipe UNION
        SELECT 'Sudah Approved Atasan' UNION 
        SELECT 'Dalam Proses Pengerjaan IT' UNION 
        SELECT 'Sudah Selesai Tapi Belum Ditutup User' UNION 
        SELECT 'Ditutup'
    ) tipe
)standard 
ON standard.tipe = requests.status

ORDER BY requests.tanggal_permintaan

4 个答案:

答案 0 :(得分：2)

要获得该输出，您需要使用CROSS JOIN表中每个不同的日期来创建tipe值的request：

SELECT r.date, t.tipe
FROM (SELECT 'Menunggu Approve Atasan' AS tipe UNION
      SELECT 'Sudah Approved Atasan' UNION 
      SELECT 'Dalam Proses Pengerjaan IT' UNION 
      SELECT 'Sudah Selesai Tapi Belum Ditutup User' UNION 
      SELECT 'Ditutup'
     ) t
CROSS JOIN 
    (SELECT DISTINCT DATE(tanggal_permintaan) AS date 
     FROM request
     WHERE YEAR(tanggal_permintaan) = 2018) r

这将为您提供一个如下表：

| date               | tipe                                  |
+--------------------+---------------------------------------+
| 2018-01-02         | Menunggu Approve Atasan               |
| 2018-01-02         | Sudah Approved Atasan                 |
| 2018-01-02         | Dalam Proses Pengerjaan IT            |
| 2018-01-02         | Sudah Selesai Tapi Belum Ditutup User |
| 2018-01-02         | Ditutup                               |
| 2018-01-03         | Menunggu Approve Atasan               |
| 2018-01-03         | Sudah Approved Atasan                 |
| 2018-01-03         | Dalam Proses Pengerjaan IT            |
| 2018-01-03         | Sudah Selesai Tapi Belum Ditutup User |
| 2018-01-03         | Ditutup                               |

然后您可以LEFT JOIN将其requests表中获取所需的结果：

SELECT s.date, s.tipe, COUNT(*) AS total
FROM (SELECT r.date, t.tipe
      FROM (SELECT 'Menunggu Approve Atasan' AS tipe UNION
            SELECT 'Sudah Approved Atasan' UNION 
            SELECT 'Dalam Proses Pengerjaan IT' UNION 
            SELECT 'Sudah Selesai Tapi Belum Ditutup User' UNION 
            SELECT 'Ditutup'
           ) t
      CROSS JOIN 
          (SELECT DISTINCT DATE(tanggal_permintaan) AS date 
           FROM request
           WHERE YEAR(tanggal_permintaan) = 2018) r
     ) s
LEFT JOIN request rq ON rq.status = s.tipe
GROUP BY s.date, s.tipe

答案 1 :(得分：1)

  • 我们可以在Derived表中获得2018年的所有唯一日期。数据上的Year()函数不允许它使用索引。我已将其更改为使用Between..，以便它可以使用索引。
  • 类似地，在单独的“派生表”中获取所有唯一状态。
    • 在它们之间
  • Cross Join以获得所有可能的组合。
  • 现在，将此all_combinations和left join用于主要查询，启用 status和date，以获取所需的结果集。

尝试以下操作：

SELECT all_combinations.tanggal_permintaan, 
       all_combinations.tipe, 
       COALESCE(requests.total, 0) AS total 
FROM 
(
  SELECT all_tanggal_permintaan.tanggal_permintaan, 
         all_tipe.tipe 
  FROM 
  (SELECT DISTINCT DATE(tanggal_permintaan) as tanggal_permintaan
   FROM request 
   WHERE tanggal_permintaan BETWEEN '2018-01-01 00:00:00' AND 
                                    '2018-12-31 23:59:59'
  ) AS all_tanggal_permintaan 

  CROSS JOIN 

  (SELECT 'Menunggu Approve Atasan' AS tipe UNION ALL
   SELECT 'Sudah Approved Atasan' UNION ALL
   SELECT 'Dalam Proses Pengerjaan IT' UNION ALL
   SELECT 'Sudah Selesai Tapi Belum Ditutup User' UNION ALL
   SELECT 'Ditutup'
  ) AS all_tipe
) AS all_combinations 

LEFT OUTER JOIN

(
  SELECT
    DATE(tanggal_permintaan) as tanggal_permintaan,
    status,
    COUNT(*) AS total 
  FROM request
  WHERE tanggal_permintaan BETWEEN '2018-01-01 00:00:00' AND 
                                   '2018-12-31 23:59:59'
  GROUP BY DATE(tanggal_permintaan), status
) AS requests 
ON requests.tanggal_permintaan = all_combinations.tanggal_permintaan AND 
   requests.status = all_combinations.tipe

ORDER BY all_combinations.tanggal_permintaan

答案 2 :(得分：1)

这是一个猜测，但似乎您需要一个状态值列表，这让我感到惊讶，因为它尚未在表中，但这可能有用：

SELECT
    g.tanggal_permintaan, g.total, s.status
FROM (
    SELECT DISTINCT
        request.status
    FROM request
) s
LEFT JOIN (
        SELECT
            DATE(request.tanggal_permintaan) AS tanggal_permintaan
           ,request.status
           ,COUNT(*)                         AS total
        FROM request
        WHERE request.tanggal_permintaan) >= '20180101'
          AND request.tanggal_permintaan) <  '20190101'
        GROUP BY
            DATE(request.tanggal_permintaan)
           ,request.status
    ) g
    ON s.status = g.status

答案 3 :(得分：0)

感谢兄弟@Nick和@Used_By_Already

这是我的最终查询。 我创建了一个基于id的唯一concat，以通过左联接将它们连接起来。

SELECT standard.*, transactions.* FROM (
    SELECT CONCAT(r.date, '#', t.tipe) as 'id', r.date AS tanggal_permintaan, t.tipe
    FROM (
         SELECT DISTINCT
             request.status AS tipe
         FROM request
    ) t
    CROSS JOIN (
        SELECT DISTINCT DATE(tanggal_permintaan) AS date 
        FROM request
        WHERE YEAR(request.tanggal_permintaan) = '2018' 
    ) r

) AS standard

LEFT JOIN (
    SELECT CONCAT(requests.tanggal_permintaan, '#' , requests.status) AS 'id', requests.tanggal_permintaan, requests.status, requests.total 
    FROM (
        SELECT
            DATE(request.tanggal_permintaan) as tanggal_permintaan,
            request.`status`,
            COUNT(*) as total
        FROM request
        WHERE YEAR(request.tanggal_permintaan) = '2018'
        GROUP BY DATE(request.tanggal_permintaan), request.status
    ) requests
) AS transactions
ON transactions.id = standard.id

ORDER BY standard.id
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