SQL 查询具有带有二进制指示符和时间差的新列

Posted 2023-03-31

【中文标题】SQL 查询具有带有二进制指示符和时间差的新列

【英文标题】：SQL query to have new columns with binary indicator and time difference

【发布时间】：2018-08-01 03:28:17

【问题描述】：

我正在尝试从 table1 编写 sql 查询以获得类似表 2 的结果，但未能成功。 任何建议将不胜感激。

我有一个如下表：

+--------+--------------+----------+---------------+-----------------------+
| UserID | account_type |  Region  | account_Level | Level_assessment_date |
+--------+--------------+----------+---------------+-----------------------+
|   1234 | A            | west268  | Normal        | Jul-17                |
|   1234 | A            | west268  | platinum      | Jul-18                |
|   2342 | B            | north234 | Normal        | May-15                |
|   3565 | A            | west268  | Normal        | Jun-14                |
|   3565 | B            | west268  | Normal        | May-17                |
|   5678 | A            | west268  | platinum      | Sep-15                |
|   6689 | A            | north234 | Normal        | Oct-16                |
|   6689 | B            | north234 | platinum      | Jan-18                |
+--------+--------------+----------+---------------+-----------------------+

我想查询 table1 以获取 table2，如下所示：

+--------+--------------+----------+---------------+-----------------------+----------------------+----------------------+
| UserID | account_type |  Region  | account_Level | Level_assessment_date | upgraded_to_platinum | Time_to_upgrade_DAYS |
+--------+--------------+----------+---------------+-----------------------+----------------------+----------------------+
|   1234 | A            | west268  | Normal        | Jul-17                | Y                    | 365                  |
|   2342 | B            | north234 | Normal        | May-15                | N                    | N/A                  |
|   3565 | A            | west268  | Normal        | Jun-14                | N                    | N/A                  |
|   3565 | B            | west268  | Normal        | May-17                | N                    | N/A                  |
|   6689 | A            | north234 | Normal        | Oct-16                | N                    | N/A                  |
+--------+--------------+----------+---------------+-----------------------+----------------------+----------------------+

正常的每个用户帐户级别的记录 ，带有二进制（是/否）列upgrade-to-platinum，表示帐户级别是否已从普通升级到白金。 和列 Time_to_upgrade_DAYS 从正常到白金所需的天数（如果有）

【问题讨论】：

你的 rdbms 是什么？ Sql Server、postgres、oracle？

为什么5678 没有出现在结果中？为什么每个 account_level 都有 Normal？

非常感谢您的帮助。我使用 sql 服务器。 5678 不在结果中，因为我只需要找到 account_level "Normal" 的结果，所以结果中不会出现 "platinum" account_level

【参考方案1】：

如果你有 sql-server 你可以试试这个。

DECLARE @table1  TABLE ( UserID INT, account_type VARCHAR(10),  Region  VARCHAR(10), account_Level VARCHAR(10), Level_assessment_date VARCHAR(10))
INSERT INTO @table1  VALUES
(1234, 'A', 'west268 ', 'Normal  ', 'Jul-17'),
(1234, 'A', 'west268 ', 'platinum', 'Jul-18'),
(2342, 'B', 'north234', 'Normal  ', 'May-15'),
(3565, 'A', 'west268 ', 'Normal  ', 'Jun-14'),
(3565, 'B', 'west268 ', 'Normal  ', 'May-17'),
(5678, 'A', 'west268 ', 'platinum', 'Sep-15'),
(6689, 'A', 'north234', 'Normal  ', 'Oct-16'),
(6689, 'B', 'north234', 'platinum', 'Jan-18')

SELECT T1.*, 
    CASE WHEN T2.UserID IS NOT NULL THEN 'Y' ELSE 'N' END upgraded_to_platinum, 
    DATEDIFF(DAY,'01-'+T1.Level_assessment_date, '01-'+T2.Level_assessment_date) Time_to_upgrade_DAYS
FROM @table1 T1
    LEFT JOIN @table1 T2 ON T1.UserID = T2.UserID and T1.account_type = T2.account_type and T2.account_Level ='platinum'
WHERE 
    T1.account_Level = 'Normal'

结果：

UserID      account_type Region     account_Level Level_assessment_date upgraded_to_platinum Time_to_upgrade_DAYS
----------- ------------ ---------- ------------- --------------------- -------------------- --------------------
1234        A            west268    Normal        Jul-17                Y                    365
2342        B            north234   Normal        May-15                N                    NULL
3565        A            west268    Normal        Jun-14                N                    NULL
3565        B            west268    Normal        May-17                N                    NULL
6689        A            north234   Normal        Oct-16                N                    NULL

【讨论】：

不客气！如果它解决了问题，您可以将其标记为答案。

【参考方案2】：

SQL DEMO

SELECT "UserID", "account_type",
       MAX("Region") as Region,
       MAX("account_Level") as account_Level,
       MAX(CASE WHEN "account_Level" = 'platinum' 
                THEN 'yes'
                ELSE 'no'
           END) as upgraded_to_platinum,
       MAX(CASE WHEN "account_Level" = 'platinum' THEN "Level_assessment_date" END) -
       MAX(CASE WHEN "account_Level" <> 'platinum' THEN "Level_assessment_date" END) as Time_to_upgrade_DAYS
FROM Table1
GROUP BY "UserID", "account_type"
ORDER BY "UserID", "account_type"

输出

| UserID | account_type |   region | account_level | upgraded_to_platinum |                             time_to_upgrade_days |
|--------|--------------|----------|---------------|----------------------|--------------------------------------------------|
|   1234 |            A |  west268 |      platinum |                  yes | 0 years 0 mons 365 days 0 hours 0 mins 0.00 secs |
|   2342 |            B | north234 |        Normal |                   no |                                           (null) |
|   3565 |            A |  west268 |        Normal |                   no |                                           (null) |
|   3565 |            B |  west268 |        Normal |                   no |                                           (null) |
|   5678 |            A |  west268 |      platinum |                  yes |                                           (null) |
|   6689 |            A | north234 |        Normal |                   no |                                           (null) |
|   6689 |            B | north234 |      platinum |                  yes |                                           (null) |