在STATION中以最短和最長的城市名稱查詢兩個城市，

Question

查詢：用最短和最長的城市名稱以及它們各自的長度（即名稱中的字符數）查詢STATION中的兩個城市。如果有多個最小或最大的城市，請選擇按字母順序排列的第一個城市。

樣品輸入：

比方說那個城市只有四個條目：DEF, ABC, PQRS and WXY

樣本輸出：

ABC 3 
PQRS 4 

Answer 1

使用UNION試試這個：

SELECT MIN(city), LENGTH(city) 
FROM Station 
WHERE LENGTH(city) = 
(SELECT MIN(LENGTH(city)) 
FROM Station) 
UNION 
SELECT MIN(city), LENGTH(city) 
FROM Station 
WHERE LENGTH(city) = 
(SELECT MAX(LENGTH(city)) 
FROM Station) 

當然，我的假設是你的表名是Statio n和列名是City。可以看看下面的只按字母順序選擇第一個記錄的相關帖子：

Return only the first alphabetical result of SELECT

Answer 2

這裏是另一種方式來做到這一點使用總是得心應手row_number解析函數：

with cte as (
    select city, 
     length(city) as len, 
     row_number() over (order by length(city), city) as smallest_rn, 
     row_number() over (order by length(city) desc, city) as largest_rn 
    from station 
) 
select city, len 
    from cte 
where smallest_rn = 1 
union all 
select city, len 
    from cte 
where largest_rn = 1 

Answer 3

select min(city), len 
    from (
     select city, length(city) len, 
       max(length(city)) over() maxlen, 
       min(length(city)) over() minlen 
      from station 
     ) 
where len in(minlen,maxlen) 
group by len 

子查詢得到城市名單和它的長度。同時"window functions"min/max over()對於set（table）中的所有行獲得最小和最大長度。主要查詢過濾器只有城市的長度是最小/最大。 min(city)與羣組len按字母順序給出結果的名字。

Answer 4

對於MS SQL服務器：

Declare @Small int 
Declare @Large int 
select @Small = Min(Len(City)) from Station 
select @Large = Max(Len(City)) from Station 
select Top 1 City as SmallestCityName,Len(City) as Minimumlength from Station where Len(City) = @Small Order by City Asc 
select Top 1 City as LargestCityName,Len(City) as MaximumLength from Station where Len(City) = @Large Order by City Asc 

對於Oracle服務器：

select * from(select distinct city,length(city) from station order by length(city) asc,city asc) where rownum=1 union 
select * from(select distinct city,length(city) from station order by length(city) desc,city desc) where rownum=1; 

Answer 5

試試這個:)

MySQL代碼....簡單的

select CITY,LENGTH(CITY) from STATION order by Length(CITY) asc, CITY limit 1; 
select CITY,LENGTH(CITY) from STATION order by Length(CITY) desc, CITY limit 1; 

Answer 6

(select CITY, 
     char_length(CITY) as len_city 
    from STATION 
    where char_length(CITY)=(select char_length(CITY) 
          from STATION 
          order by char_length(CITY) LIMIT 1) 
    Order by CITY LIMIT 1) 
UNION ALL 
(select CITY, 
     char_length(CITY) as len_city 
    from STATION 
    where char_length(CITY)=(select char_length(CITY) 
          from STATION 
          order by char_length(CITY) DESC LIMIT 1) 
    Order by CITY DESC LIMIT 1) 
    ORDER BY char_length(CITY); 

Answer 7

SELECT TOP 1 CITY,LEN(CITY) from STATION 
    Where LEN(CITY)=(Select min(len(CITY)) from STATION) order by CITY 
SELECT TOP 1 CITY,LEN(CITY) from STATION 
    Where LEN(CITY)=(Select MAX(len(CITY)) from STATION)