如何改善这4个自我加入？

Question

假设我有下面的示例数据集：

emplid | Citizenship | 
100001 | USA   | 
100001 | CAN   | 
100001 | CHN   | 
100002 | USA   | 
100002 | CHN   | 
100003 | USA   | 

我想安排它要为每个员工的国籍一行。我们可以假设一名员工拥有多达四个国籍。输出应该是这样的：

emplid | Citizeship_1 | Citizenship_2 | Citizenship_3 
100001 | USA   | CHN   | CAN 
100002 | USA   | CHN   | 
100003 | USA   |    | 

唯一可行的解​​决方案我已经能够做到这一点是：由于数据集的增长和增长这变得越来越低效

SELECT e.emplid, MAX(e.citizenship) AS citizenship1, 
       MAX(e1.citizenship) AS citizenship2, 
       MAX(e2.citizenship) AS citizenship3, 
       MAX(e3.citizenship) AS citizenship4 
FROM employee e 
LEFT JOIN employee e1 ON e1.emplid = e.emplid AND e1.citizenship < e.citizenship 
LEFT JOIN employee e2 ON e2.emplid = e1.emplid AND e2.citizenship < e1.citizenship 
LEFT JOIN employee e3 ON e3.emplid = e2.emplid AND e3.citizenship < e2.citizenship 
GROUP BY e.emplid 

，但我找不到重写此查询的方法。

Answer 1

为什么不把公民资格列入清单？

select e.emplid, group_concat(citizenship) as citizenships 
from employee e 
group by e.emplid; 

如果你想有四个单独的列，你可以这样做：

select e.emplid, 
     substring_index(group_concat(citizenship), ',', 1) as c1, 
     (case when count(*) >= 2 
      then substring_index(substring_index(group_concat(citizenship), ',', 2), ',', -1) 
     end) as c2, 
     (case when count(*) >= 3 
      then substring_index(substring_index(group_concat(citizenship), ',', 3), ',', -1) 
     end) as c3, 
     (case when count(*) >= 4 
      then substring_index(substring_index(group_concat(citizenship), ',', 4), ',', -1) 
     end) as c4 
from employee e 
group by e.emplid; 

Answer 2

该解决方案按字母顺序排的每个员工的国籍，然后将结果付诸相应的列。

SELECT 
    emplid, 
    MAX(CASE WHEN R = 1 THEN Citizenship ELSE NULL END) AS Citizeship_1, 
    MAX(CASE WHEN R = 2 THEN Citizenship ELSE NULL END) AS Citizeship_2, 
    MAX(CASE WHEN R = 3 THEN Citizenship ELSE NULL END) AS Citizeship_3 
FROM  
    (SELECT emplid,Citizenship,RANK() OVER(PARTITION BY emplid ORDER BY Citizenship) AS R FROM @T) AS DATA 
GROUP BY 
    emplid