Len中的Column查詢

Question

我是SQL新手，因此問題。

因此，我用下面的表，以Id，城市，州名爲站。

我需要在STATION中以最短和最長的CITY名稱以及它們各自的長度（即名稱中的字符數）來查詢兩個城市。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。

有人可以幫我開始做這件事。我試過函數city，但它似乎不工作。

Answer 1

市最短名稱：

SELECT City, CHAR_LENGTH(City) AS len 
FROM STATION 
ORDER BY CHAR_LENGTH(City) ASC, City 
LIMIT 1 

市最長的名字：

SELECT City, CHAR_LENGTH(City) AS len 
FROM STATION 
ORDER BY CHAR_LENGTH(City) DESC, City 
LIMIT 1 

你可以將它們組合成一個單一的語句UNION ALL：

SELECT * 
FROM (
    SELECT City, CHAR_LENGTH(City) AS len 
    FROM STATION 
    ORDER BY CHAR_LENGTH(City) ASC, City 
    LIMIT 1 
) shortest 
UNION ALL (
    SELECT City, CHAR_LENGTH(City) AS len 
    FROM STATION 
    ORDER BY CHAR_LENGTH(City) DESC, City 
    LIMIT 1 
) longest 

Answer 2

這裏是一個單一的查詢，你可以嘗試：

SELECT City, CHAR_LENGTH(City) AS length, 'max char length' AS description 
FROM yourTable 
WHERE CHAR_LENGTH(City) = (SELECT MIN(CHAR_LENGTH(City)) FROM yourTable) 
ORDER BY City 
LIMIT 1 
UNION 
SELECT City, CHAR_LENGTH(City) AS length, 'min char length' AS description 
FROM yourTable 
WHERE CHAR_LENGTH(City) = (SELECT MAX(CHAR_LENGTH(City)) FROM yourTable) 
ORDER BY City 
LIMIT 1 

Answer 3

這裏是我的裂紋它。這是未經測試的，但它會給你一些開始。 （編輯：現在它已經過測試，它的工作原理可能會更好，但這將是使用子查詢的替代方案）。

SELECT 
    (SELECT City 
    FROM table 
    ORDER BY CHAR_LENGTH(City) DESC, City 
    LIMIT 1) AS longest, 
    (SELECT CHAR_LENGTH(City) 
    FROM table 
    ORDER BY CHAR_LENGTH(City) DESC, City 
    LIMIT 1) AS longest_length, 
    (SELECT City 
    FROM table 
    ORDER BY CHAR_LENGTH(City) ASC, City 
    LIMIT 1) AS shortest, 
    (SELECT CHAR_LENGTH(City) 
    FROM table 
    ORDER BY CHAR_LENGTH(City) ASC, City 
    LIMIT 1) AS shortest_length 
FROM table 
LIMIT 1 

Answer 4

嘿得到了黑客排名:-P

同樣的問題，這是解決方案：

select * from 
(select city, length(city) from station where length(city) = (select min(length(city)) from station) order by 1 limit 1) as tmp1 
union all 
select * from 
(select city, length(city) from station where length(city) = (select max(length(city)) from station) order by 1 limit 1) as tmp2 

我們不能子句UNION子句中使用限制和秩序。因此我們需要從查詢和應用UNION子句中取得結果的別名。

希望這將有助於:-)

Answer 5

(select CITY,length(CITY) 
from STATION 
where length(CITY)=(select min(length(CITY)) from STATION) 
order by city 
LIMIT 1) 
union 
(select CITY,length(CITY) 
from STATION 
where length(CITY)=(select max(length(CITY)) from STATION) 
order by city 
LIMIT 1); 

Answer 6

(select CITY,length(CITY) 
from STATION 
where length(CITY)=(select min(length(CITY)) from STATION) 
order by city 
LIMIT 1) 
union 
(select CITY,length(CITY) 
from STATION 
where length(CITY)=(select max(length(CITY)) from STATION) 
order by city 
LIMIT 1) 

Answer 7

這也可行 -

select first(CITY) ,len(first(CITY)) from STATION where CITY in (select CITY from STATION where len(CITY) in (SELECT MAX(LEN(CITY)) FROM STATION) order by CITY asc) 
UNION 
select first(CITY) ,len(first(CITY)) from STATION where CITY in (select CITY from STATION where len(CITY) in (SELECT MIN(LEN(CITY)) FROM STATION) order by CITY asc); 

Answer 8

SELECT City, CHAR_LENGTH(City) 
FROM STATION 
ORDER BY CHAR_LENGTH(City) ASC, City 
LIMIT 1; 

SELECT City, CHAR_LENGTH(City) 
FROM STATION 
ORDER BY CHAR_LENGTH(City) DESC, City 
LIMIT 1; 

試試上面的解決方案，我估計應該解決您的問題。

Answer 9

這爲我工作：

（選擇城市，CHAR_LENGTH（CITY）FROM STATION ORDER BY CHAR_LENGTH（CITY）ASC LIMIT 1）UNION（選擇城市，CHAR_LENGTH（CITY）FROM STATION ORDER BY CHAR_LENGTH（CITY） DESC LIMIT 1）