Case语句的SQL问题和连接多个表

Question

这是两个表的示例（请参阅 SQLFiddle），我尝试创建一个 select 语句，其中一个表是查找表（Facility），另一个是数据表（WOENTITY），我想从中选择获取 facititynames查找并且每当它找到多个设施ID条目时，只需将文本写入“多个位置”，或者如果它找到空或零然后写入“未知”，否则只需写入设施名称，我试图做一个案例以用单个条目替换多个计数。任何帮助表示赞赏。

链接到 SQL Fiddle：Click Here

这是表格外观的屏幕截图，Table WOEntity 和 Table Facility Image showing the table schema and expected end result

我首先想到从 WOEntity 表中获取多个位置 ID 的计数，然后创建一个 select Case 语句但没有用，这里是草稿代码。

select e.WOID, count(e.ENTITYUID) as CNT, 
    (case   
            when e.ENTITYUID <> '0' and count(e.ENTITYUID) <=1 then
                (Select FACILITYNAME from FACILITY l where e.ENTITYUID = 
                  l.FACILITYID)
            when count(e.ENTITYUID) > 2 then 'MULTIPLE FACILITIES'
            else 'UNKNOWN LOCATIONS'
    end) as facilityName
 From WOENTITY e
group by e.WOID order by WOID;

Answer 1

我个人会使用 JOIN 而不是相关子查询来执行此操作，然后在 case 表达式中使用 COUNT：

SELECT  e.WOID,
        CASE WHEN COUNT(f.FACILITYID) = 0 THEN 'UNKNOWN LOCATIONS'
            WHEN COUNT(f.FACILITYID) > 1 THEN 'MULTIPLE FACILITIES'
            ELSE MAX(f.FACILITYNAME)
        END AS facilityName
FROM    WOENTITY AS e
        LEFT JOIN FACILITY AS f
            ON f.FACILITYID = e.ENTITYUID
GROUP BY e.WOID
ORDER BY e.WOID;

SQL Fiddle

由于您处理 JOIN 匹配的多条记录或零条记录的情况，当您到达 ELSE 时，您只有一条记录，因此尽管您使用的是 MAX 聚合函数，您只有一个输入行，所以除了允许您选择设施名称而不按其分组之外，这并没有真正做任何事情。

编辑

我刚刚注意到加拿大应该显示为多个位置，即使它们中只有一个存在于设施表中，在这种情况下，您需要稍微更改大小写表达式，将 COUNT(f.FACILITYID) 更改为 COUNT(*) 第二个谓词：

SELECT  e.WOID,
        CASE WHEN COUNT(f.FACILITYID) = 0 THEN 'UNKNOWN LOCATIONS'
            WHEN COUNT(*) > 1 THEN 'MULTIPLE FACILITIES'
            ELSE MAX(f.FACILITYNAME)
        END AS facilityName
FROM    WOENTITY AS e
        LEFT JOIN FACILITY AS f
            ON f.FACILITYID = e.ENTITYUID
GROUP BY e.WOID
ORDER BY e.WOID;

SQL Fiddle

Answer 2

正如声明对您所说，您错过了 params 的一个 gorup ... 类似：

select e.WOID, count(e.ENTITYUID) as CNT, 
    (case   
            when e.ENTITYUID <> '0' and count(e.ENTITYUID) <=1 then
                (Select FACILITYNAME from FACILITY l where e.ENTITYUID = 
                  l.FACILITYID) 
            when count(e.ENTITYUID) > 2 then 'MULTIPLE FACILITIES'
            else 'UNKNOWN LOCATIONS'
    end) as facilityName
 From WOENTITY e
group by e.WOID,e.ENTITYUID order by WOID;

Answer 3

请试试这个

Select FACILITYID,
    CASE WHEN (SELECT COUNT(*) 
               FROM  WOEnTITY 
               WHERE WOEnTITY.ENTITYUID = FACILITYID)  > 0 
         THEN 'MULTIPLE LOCATIONS' 
         ELSE '' 
    END
FROM FACILITY

Answer 4

您可以如下查询：

Select Row_Number() over(order by Id) as Id, Woid, Case when FACILITYID is null then 'UNKNOWN LOCATIONS'
               when RowCnt > 1 then 'MULTIPLE OFFICES' else FACILITYNAME end as FacilityName,
               RowCnt as CountFacilities 
               from (
    Select *, RowCnt = count(EntityUid) over(partition by woid),
    RowN = Row_Number() over(Partition by woid order by Id)
    from WOENTITY we
    left join Facility f
    on we.ENTITYUID = f.FACILITYID
) a
Where a.RowN = 1

输出如下：

+----+---------+-------------------+-----------------+
| Id |  Woid   |   FacilityName    | CountFacilities |
+----+---------+-------------------+-----------------+
|  1 | Canada  | MULTIPLE OFFICES  |               2 |
|  2 | Germany | GREEN             |               1 |
|  3 | India   | GREEN             |               1 |
|  4 | UK      | RED               |               1 |
|  5 | US      | MULTIPLE OFFICES  |               3 |
|  6 | JAPAN   | UNKNOWN LOCATIONS |               1 |
+----+---------+-------------------+-----------------+

Answer 5

这是对@GarethD 答案的修改，考虑了类似于“加拿大”的情况

  SELECT  e.WOID,
        CASE WHEN COUNT(f.FACILITYID) = 0 THEN 'UNKNOWN LOCATIONS'
            WHEN COUNT(f.FACILITYID) >0 and  COUNT(coalesce(f.FACILITYID,'x'))>COUNT(f.FACILITYID)  then 'MULTIPLE OFFICES'
            WHEN COUNT(f.FACILITYID) > 1 then 'MULTIPLE OFFICES'
            ELSE MAX(f.FACILITYNAME)
        END AS facilityName
FROM    WOENTITY AS e
        LEFT JOIN FACILITY AS f
            ON f.FACILITYID = e.ENTITYUID
GROUP BY e.WOID
ORDER BY e.WOID;