Question

我正在寻找一个SQL查询（甚至更好的LINQ查询）来删除已经取消休假的人，即删除所有具有相同NAME并且START和END相同的记录，且DAYS_TAKEN值仅在符号上有所不同

如何从中获取

NAME    |DAYS_TAKEN |START      |END        |UNIQUE_LEAVE_ID    
--------|-----------|-----------|-----------|-----------
Alice   |  2        | 1 June    | 3 June    | 1 --remove because cancelled
Alice   | -2        | 1 June    | 3 June    | 2 --cancelled
Alice   |  3        | 5 June    | 8 June    | 3 --keep
Bob     | 10        | 4 June    | 14 June   | 4 --keep
Charles | 12        | 2 June    | 14 June   | 5 --remove because cancelled
Charles | -12       | 2 June    | 14 June   | 6 --cancelled
David   | 5         | 3 June    | 8 June    | 7 --keep

对此？

NAME    |DAYS_TAKEN |START      |END        |UNIQUE_LEAVE_ID    
--------|-----------|-----------|-----------|-----------
Alice   |  3        | 5 June    | 8 June    | 3 --keep
Bob     | 10        | 4 June    | 14 June   | 4 --keep
David   | 5         | 3 June    | 8 June    | 7 --keep

我尝试了什么

Query1查找所有已取消的记录（不确定这是否正确）

SELECT L1.UNIQUE_LEAVE_ID 
FROM LEAVE L1 
INNER JOIN LEAVE L2 ON L2.DAYS_TAKEN > 0 AND ABS(L1.DAYS_TAKEN) = L2.DAYS_TAKEN AND L1.NAME= L2.NAME AND L1.START = L2.START AND L1.END = L2.END
WHERE L1.DAYS_TAKEN < 0

然后我在内部选择中使用Query1两次，如此

SELECT L.* FROM LEAVE L WHERE 
L.UNIQUE_LEAVE_ID NOT IN (Query1)
AND L.UNIQUE_LEAVE_ID NOT IN (Query1)

有没有办法只使用内部查询一次？

（这是一个Oracle数据库，从.NET / C＃调用）

Answer 1

您可以使用如下查询：

SELECT NAME, START, END  
FROM LEAVE
GROUP BY NAME, START, END
HAVING SUM(DAYS_TAKEN) = 0

为了获取已取消的NAME, START, END个群组（假设取消记录的DAYS_TAKEN取消了初始记录的天数。）

<强>输出：

NAME    |START      |END        
--------|-----------|----------
Alice   | 1 June    | 3 June
Charles | 2 June    | 14 June

使用上述查询作为派生表，您可以获得与“已取消”组无关的记录：

SELECT L1.NAME, L1.DAYS_TAKEN, L1.START, L1.END, L1.UNIQUE_LEAVE_ID  
FROM LEAVE L1
LEFT JOIN (
  SELECT NAME, START, END  
  FROM LEAVE
  GROUP BY NAME, START, END
  HAVING SUM(DAYS_TAKEN) = 0
) L2 ON L1.NAME = L2.NAME AND L1.START = L2.START AND L1.END = L2.END
WHERE L2.NAME IS NULL

<强>输出：

NAME    |DAYS_TAKEN |START      |END        |UNIQUE_LEAVE_ID    
--------|-----------|-----------|-----------|-----------
Alice   | 3         | 5 June    | 8 June    | 3 
Bob     | 10        | 4 June    | 14 June   | 4 
David   | 5         | 3 June    | 8 June    | 7

Answer 2

您可以使用not exists：

select l.*
from leave l
where not exists (select 1
                  from leave l2
                  where l2.name = l.name and l2.start = l.start and
                        l2.end = l.name and l2.days_taken = - l.days_taken
                 );

此查询可以利用leave(name, start, end, days_taken)上的索引。

Answer 3

以下是SUM() OVER的变体：

SELECT x.*
  FROM (SELECT l.*, SUM (days_taken) OVER (PARTITION BY name, "START", "END", ABS (days_taken) ORDER BY NULL) s
          FROM leave l) x
 WHERE s <> 0

如果你有Oracle 12，这会给你取消：

SELECT l.*
  FROM leave l,
       LATERAL (SELECT days_taken
                  FROM leave l2
                 WHERE l2.name = l.name 
                   AND l2."START" = l."START" 
                   AND l2."END" = l."END" 
                   AND l2.days_taken = -l.days_taken) x

这应该保留：

SELECT l.*
  FROM leave l
       OUTER APPLY (SELECT days_taken
                      FROM leave l2
                     WHERE l2.name = l.name 
                       AND l2."START" = l."START" 
                       AND l2."END" = l."END" 
                       AND l2.days_taken = -l.days_taken) x
 WHERE x.days_taken IS NULL

关于列名称的一些内容。建议不要在Oracle SQL中使用保留字，但如果必须这样做，请使用＆＃39;＆＃34;＆＃39;喜欢这里。

Answer 4

我使用Giorgos的答案来提出这个Linq解决方案。该解决方案还考虑了多次取消/休假的人。见下面的爱丽丝和埃德加。

示例数据

  int id = 0;
  List<Leave> allLeave = new List<Leave>()
  {
    new Leave() { UniqueLeaveID=id++, Name="Alice", Start=new DateTime(2016,6,1), End=new DateTime(2016,6,3), Taken=-2 },
    new Leave() { UniqueLeaveID=id++,Name="Alice", Start=new DateTime(2016,6,1), End=new DateTime(2016,6,3), Taken=2 },
    new Leave() { UniqueLeaveID=id++, Name="Alice", Start=new DateTime(2016,6,1), End=new DateTime(2016,6,3), Taken=2 },
    new Leave() { UniqueLeaveID=id++,Name="Alice", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,5), Taken=3 },

    new Leave() { UniqueLeaveID=id++,Name="Bob", Start=new DateTime(2016,6,4), End=new DateTime(2016,6,14), Taken=10 },

    new Leave() { UniqueLeaveID=id++,Name="Charles", Start=new DateTime(2016,6,2), End=new DateTime(2016,6,14), Taken=12 },
    new Leave() { UniqueLeaveID=id++,Name="Charles", Start=new DateTime(2016,6,2), End=new DateTime(2016,6,14), Taken=-12 },

    new Leave() { UniqueLeaveID=id++,Name="David", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,8), Taken=5 },

            new Leave() { UniqueLeaveID=id++,Name="Edgar", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,8), Taken=5 },
    new Leave() { UniqueLeaveID=id++,Name="Edgar", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,8), Taken=5 },
    new Leave() { UniqueLeaveID=id++,Name="Edgar", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,8), Taken=5 },
    new Leave() { UniqueLeaveID=id++,Name="Edgar", Start=new DateTime(2016,6,3), End=new DateTime(2016,6,8), Taken=5 }

  };

Linq Query（注意Oracle版本11与12）

 var filteredLeave = allLeave
 .GroupBy(a => new { a.Name, a.Start, a.End })
 .Select(a => new { Group = a.OrderByDescending(b=>b.Taken), Count = a.Count() })
 .Where(a => a.Count % 2 != 0)
 .Select(a => a.Group.First());

＆＃34; OrderByDescending＆＃34;确保只返回正面的日子。

Oracle SQL

SELECT
*
FROM
(
    SELECT 
    L1.NAME, L1.START, L1.END, MAX(TAKEN) AS TAKEN, COUNT(*) AS CNT
    FROM LEAVE L1
    GROUP BY L1.NAME, L1.START, L1.END
) L2
WHERE MOD(L2.CNT,2)<>0 -- replace MOD with % for Microsoft SQL

条件＆＃34; WHERE MOD（L2.CNT，2）＆lt;＆gt; 0＆＃34; （或在Linq＆＃34; a.Count％2！= 0＆＃34;）仅返回申请一次或奇数次的人（例如申请 - 取消 - 申请）。但申请 - 取消 - 申请 - 取消的人会被过滤掉。

删除所有符号相反的记录

4 个答案: