SQL：如何获取列中不同值的总数？

Question

SQL初学者在这里。当前正在处理mySQL和Postgre SQL的问题。

我想获取每个订单优先级的总计（Not_Specified，Low，Medium，High，Critical）对于每个州。

例如，我想为得克萨斯州创建一列，为每个订单优先级类别分配一个数字，然后为下一个州分配一个数字，依此类推。每个订单优先级在每个状态下都有自己的计数栏。

这是我当前在下面的查询。我可以使用子查询还是需要使用窗口函数？

SELECT 
    Customer_ID, City, State_or_Province, Order_Date, Order_Priority, 
    ROW_NUMBER() OVER(ORDER BY City ASC, State_or_Province ASC) AS Row_N,
    COUNT(Order_Priority) OVER (Partition BY State_or_Province) AS State_Total_count

FROM SuperStore_Main 

Answer 1

您似乎正在寻找条件聚合。

在MySQL中：

select
    state_or_province,
    sum(order_priority = 'Not_Specified') cnt_not_specified,
    sum(order_priority = 'Low')           cnt_low
    sum(order_priority = 'Medium')        cnt_medium
    sum(order_priority = 'High')          cnt_not_high
    sum(order_priority = 'Critical')      cnt_critical
from superstore_main
group by state_or_province

在Postgres：

select
    state_or_province,
    count(*) filter(where order_priority = 'Not_Specified') cnt_not_specified,
    count(*) filter(where order_priority = 'Low')           cnt_low
    count(*) filter(where order_priority = 'Medium')        cnt_medium
    count(*) filter(where order_priority = 'High')          cnt_not_high
    count(*) filter(where order_priority = 'Critical')      cnt_critical
from superstore_main
group by state_or_province

Answer 2

此PostgreSQL查询分解了状态和顺序优先级的每种组合的记录数：

  SELECT State_or_Province
       , Order_Priority
       , COUNT(*) tally
    FROM SuperStore_Main 
GROUP BY State_or_Province
       , Order_Priority
       ;