Question

我会尝试解释我想要快速实现的目标，因为我不知道如何解释它！

我们这里有一张表格，显示所有员工的所有工作经历，我想要当前帖子的“Start_Date”（“Current_Flag”='Y'）。除此之外，我想要之前的帖子的“End_Date”（将按当前标记过滤，按结束日期排序，然后抓住顶部日期）

无论如何，这是我的代码：

SELECT "Gc_Staff_Number",
       "Start_Date",
       (SELECT "End_Date"
        FROM   "Employment_History"
        WHERE  "Current_Flag" != 'Y'
               AND ROWNUM = 1
               AND "Employee_Number" = "Employment_History"."Employee_Number"
        ORDER  BY "End_Date" ASC)
FROM   "Employment_History"
WHERE  "Current_Flag" = 'Y'

关于如何使这项工作的任何建议都很棒，希望上面的内容有点意义 - 说实话，此刻的查询甚至无法正常工作，嗯。

（编辑：哦！我写这个是为了查询一个现有的系统......由于某种原因，桌子周围有所有愚蠢的双引号和字段名称，叹了口气！）

Answer 1

这正是分析拯救的一种场景。

鉴于此测试数据：

SQL> select * from employment_history
  2  order by Gc_Staff_Number
  3             , start_date
  4  /

GC_STAFF_NUMBER START_DAT END_DATE  C
--------------- --------- --------- -
           1111 16-OCT-09           Y
           2222 08-MAR-08 26-MAY-09 N
           2222 12-DEC-09           Y
           3333 18-MAR-07 08-MAR-08 N
           3333 01-JUL-09 21-MAR-09 N
           3333 30-JUL-10           Y

6 rows selected.

SQL>

带有分析LAG（）函数的内联视图提供了正确的答案：

SQL> select Gc_Staff_Number
  2             , start_date
  3             , prev_end_date
  4  from   (
  5      select Gc_Staff_Number
  6             , start_date
  7             , lag (end_date) over (partition by Gc_Staff_Number
  8                                    order by start_date )
  9                  as prev_end_date
 10             , current_flag
 11      from employment_history
 12  )
 13  where current_flag = 'Y'
 14  /

GC_STAFF_NUMBER START_DAT PREV_END_
--------------- --------- ---------
           1111 16-OCT-09
           2222 12-DEC-09 26-MAY-09
           3333 30-JUL-10 21-MAR-09

SQL>

内联视图对于获得正确的结果至关重要。否则，CURRENT_FLAG上的过滤器将删除先前的行。

Answer 2

我对引号感到有点困惑，但是，下面应该对你有用：

SELECT "Gc_Staff_Number",
       "Start_Date", x.end_date
FROM   "Employment_History" eh,
(SELECT "End_Date"
        FROM   "Employment_History"
        WHERE  "Current_Flag" != 'Y'
               AND ROWNUM = 1
               AND "Employee_Number" = eh.Employee_Number
        ORDER  BY "End_Date" ASC) x
WHERE  "Current_Flag" = 'Y'

Answer 3

SELECT "Gc_Staff_Number",
       "Start_Date",
       (SELECT "End_Date"
        FROM   "Employment_History"
        WHERE  "Current_Flag" != 'Y'
               AND ROWNUM = 1
               AND "Employee_Number" = "Employment_History"."Employee_Number"
        ORDER  BY "End_Date" ASC)
FROM   "Employment_History"
WHERE  "Current_Flag" = 'Y'

仅供参考，在这种情况下，ROWNUM = 1在ORDER BY之前得到评估，因此内部查询将对（最多）一条记录的总计进行排序。

如果你真的在寻找给定员工最早的end_date（current_flag＆lt;＆gt;'Y'），那么你正在寻找什么？

SELECT "Gc_Staff_Number",
       "Start_Date",
       eh.end_date
  FROM "Employment_History" eh
       LEFT OUTER JOIN -- in case the current record is the only record...
       (SELECT "Employee_Number"
             , MIN("End_Date") as end_date
          FROM "Employment_History"
         WHERE "Current_Flag" != 'Y'
         GROUP BY "Employee_Number" 
       ) emp_end_date
          ON eh."Employee_Number" = emp_end_date."Employee_Number"
 WHERE eh."Current_Flag" = 'Y'

Answer 4

SELECT eh."Gc_Staff_Number",
       eh."Start_Date",
       MAX(eh2."End_Date") AS "End_Date"
FROM   "Employment_History" eh
LEFT JOIN  "Employment_History" eh2
ON eh."Employee_Number" = eh2."Employee_Number" and eh2."Current_Flag" != 'Y'
WHERE  eh."Current_Flag" = 'Y' 
GROUP BY eh."Gc_Staff_Number",
       eh."Start_Date

Answer 5

我将LAG function用于：

SELECT eh.gc_staff_number,
       eh.start_date,
       LAG(eh.end_date) OVER (PARTITION BY eh.gc_staff_number
                                  ORDER BY eh.end_date) AS prev_end_date
  FROM EMPLOYMENT_HISTORY eh
 WHERE eh.current_flag = 'Y'

如果你想提前一行，你可以使用LEAD function。

兼容性：

据我所知，这支持9i +，但我还没有确认支持8i就像文档索赔一样。

LEAD和LAG最终是ANSI，但目前只有Oracle和PostgreSQL v8.4+支持它们。

Answer 6

基本上，你所要做的就是

select ..., (select ... from ... where ...) as ..., ..., from ... where ...

例如。您可以在任何地方插入（选择......从...到哪里），它将被相应的数据替换。

我知道其他人的例子（即使他们每个人真的很棒:)）对新手来说有点复杂（比如我：p）所以我希望这个“简单”的例子可以帮助你们中的一些人：）

Oracle SQL - 在select中选择（在同一个表中！）

6 个答案:

兼容性：