Question

首先，如果标题没有意义，我很抱歉，但下面是详细的情节。

说我有document_revision表

id   document_id  phase_id  user_id
1    1            3         1
2    1            2         1
3    1            1         1
4    2            3         2
5    2            2         2

其中phase_id是：transcribe = 3;证明= 2;并提交= 1。

我想写一个查询，我可以过滤修订记录，如果相同的用户进行了转录和证明，我将忽略证明阶段。所以输出将是：

id   document_id  phase_id  user_id
1    1            3         1
3    1            1         1
4    2            3         2

我已经挣扎了好几个小时才弄清楚这个问题，但到目前为止还没有运气。

Answer 1

假设您只想在第2阶段和第3阶段涉及user_id的任何情况下使用阶段3，那么您可以使用// // Script file 2 // Database credentials test // dbmysqlconntest = PropertiesService.getDocumentProperties(); dbmysqlconntest.setProperties({ 'mysqlUrl': 'jdbc:mysql://dbserverdev;databaseName=sample', 'mysqlUser': 'username', 'mysqlPassword': 'password'}); var keys = dbmysqlconntest.getKeys(); for (var i = 0; i < keys.length; i++) { dbmysqlconntest.getProperty(keys[i]); } var url = dbmysqlconntest.getProperty(keys[0]); var user = dbmysqlconntest.getProperty(keys[1]); var password = dbmysqlconntest.getProperty(keys[2]); function delProperty() { var userProperties = PropertiesService.getUserProperties(); userProperties.deleteProperty('mysqlUrl'); }进行此操作，例如：

ROW_NUMBER()

Answer 2

DECLARE @document_revision TABLE (
    id INT IDENTITY(1,1),
    document_id INT,
    phase_id INT,
    user_id INT
);

INSERT INTO @document_revision
(document_id, phase_id, user_id)
VALUES  
    (1, 3, 1), 
    (1, 2, 1),
    (1, 1, 1),
    (2, 3, 2),
    (2, 2, 2),
    -- To test a scenario where there is a proof and a submit with no transcribe phases and same document
    (3, 2, 3), 
    (3, 1, 3),
    -- To test a scenario where there is a transcribe and a submit with no proof phases and same document
    (4, 3, 4), 
    (4, 1, 4),
    -- To test a scenario where there is a proof and a submit with no transcribe phase (for document_id 5) but different document and same user as above 
    (5, 2, 4);

SELECT  dr.id
      , dr.document_id
      , dr.phase_id
      , dr.user_id 
FROM @document_revision AS dr
WHERE NOT EXISTS (  SELECT 1 
                    FROM @document_revision AS temp 
                    -- Same user
                    WHERE temp.user_id = dr.user_id 
                        -- Same document
                        AND temp.document_id = dr.document_id
                        -- To check if there is already a transcribe phase_id with the same user_id and document_id
                        AND temp.phase_id = 3
                        -- -- To check if there is already a proof phase_id with the same user_id and document_id
                        AND dr.phase_id = 2 )

结果：

id  document_id phase_id    user_id
1   1           3           1
3   1           1           1
4   2           3           2
6   3           2           3 
7   3           1           3
8   4           3           4
9   4           1           4
10  5           2           4

根据表的其他记录过滤表的记录

2 个答案: