SQL查詢從我現有的數據表中獲取一個州中城市最多的郵政編碼

Question

我有一個表'People'與這些列：id，fname，lname，addr，city，state，zip in Oracle 9.x中

我想人口最多的郵編每個城市每個國家

我寫此查詢：

Select City, State, Zip, count(*) 
From People 
Group By City, State 
Order By count(*) 

但它給了我一個市多行對於一個國家，贊（組成）： -

City -- State -- Zip -- Count(*) 
City0 -- ST0 -- 32111 -- 50 
City1 -- ST1 -- 11223 -- 100 
City1 -- ST1 -- 11225 -- 90 
City1 -- ST1 -- 11226 -- 50 
City2 -- ST1 -- 11255 -- 70 
City3 -- ST2 -- 55443 -- 60 

我試着像HAVING子句：具有計數（）= MAX（計數（）），但得到的錯誤消息：having子句過深或somethig嵌套（甲骨文的9.x）

我只想頂行： - City1 - ST1 - 11223 - 100 所有City1 - ST1行，其他行那裏只有一個行每個城市的每個狀態將保持不變。通緝輸出： -

City -- State -- Zip -- Count(*) 
City0 -- ST0 -- 32111 -- 50 
**City1 -- ST1 -- 11223 -- 100** 
City2 -- ST1 -- 11255 -- 70 
City3 -- ST2 -- 55443 -- 60 

我該如何做到這一點？感謝您的觀看。

==========

ANSWER FOUND吉姆·哈德森的答覆 通過修改提供的查詢一點我是正確的。最終代碼： -

select city, state, zip, counter from (
select city, state, zip, count(*) as counter from people group by city, state, zip 
) 
where counter = (
    select max(count2) from (
     select city as city1, state as state1, zip as zip1, count(*) as count2 from people group by city, state, zip ) 
    where city=city1 and state=state1 
); 

Answer 1

將您的第一個查詢視爲視圖。它給你的城市，州，郵編的計數。然後用它作爲起點。這將是容易得到最大的拉鍊人口。

例如，

select city, state, max(counter) from 
(select city, state, zip, count(*) as counter from people group by city, state, zip) 
group by city, state; 

當然，這不是很想要，因爲你想知道哪個拉鍊那是什麼。因此，舊的方式將類似於

select city, state, zip, counter from 
(select city, state, zip, count(*) as counter from people group by city, state, zip) 
where counter = (select max(count2) from (select city, state, zip, count(*) as count2 from people group by city, state, zip)); 

還有其他的方法可以做到使用分析功能的第二步。我會讓別人在這些工作。但關鍵的步驟是創建內嵌視圖，並將它作爲你進一步分析的基礎。

Answer 2

select * from (
    select city, state, zip, 
    rank() over (partition by city, state order by cnt desc) rank 
    from (
    Select City, State, Zip, count(*) cnt 
    From People 
    Group By City, State, Zip 
) 
) 
where rank = 1