Question

我想知道为什么我的代码不起作用。这个问题在此之前已被提出： Query the two cities in STATION with the shortest and longest CITY names,

和解决方案： https://github.com/chhayac/SQL-hackerrank-problems/blob/master/basic-select.md

但这两个答案都不起作用。我已粘贴下面的问题，然后是我的解决方案。谢谢你的帮助！

使用最短和最长的CITY名称查询STATION中的两个城市，以及它们各自的长度（即：名称中的字符数）。如果有多个最小或最大的城市，请选择按字母顺序排序的第一个城市。

输入格式

STATION表描述如下：

Station.jpg

其中LAT_N是北纬，LONG_W是西经。

示例输入

让我们说CITY只有四个条目：DEF，ABC，PQRS和WXY

示例输出

ABC 3
PQRS 4

解释

按字母顺序排序时，CITY名称列为ABC，DEF，PQRS和WXY，各自的长度和。名字最长的城市显然是PQRS，但有最短命名的城市可供选择;我们选择ABC，因为它首先按字母顺序排列。

请注意您可以编写两个单独的查询来获得所需的输出。它不一定是一个查询。

我的回答：

/ 按字母顺序排列的最短字符长度 /

SELECT city, LENGTH(city) as length_char
FROM station
ORDER BY LENGTH(city) ASC, city ASC
LIMIT 1;

/ 按字母顺序排列的最长字符长度 /

SELECT city, LENGTH(city) as length_char
FROM station
ORDER BY LENGTH(city) DESC
LIMIT 1;

Answer 1

您在github上的解决方案如下：

select city, length(city) from station order by length(city) DESC,city ASC fetch first row only;
select city, length(city) from station order by length(city) asc ,city asc fetch first row only;

您在这里遇到问题-没有fetch first row only这样的命令。根据数据库系统的不同，可以是top，limit或rownum-请在此处阅读更多内容-https://www.w3schools.com/sql/sql_top.asp

因此，根据系统的不同，答案也将有所不同。

Oracle

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

select * from (select city c, length(city) l
from   station
order by l asc, c asc)
where rownum = 1;

SQL Server

select top 1 city c, len(city) l
from   station
order by l desc, c asc;

select top 1 city c, len(city) l
from   station
order by l asc, c asc;

MySQL

select city c, length(city) l
from   station
order by l desc, c asc
limit 1;

select city c, length(city) l
from   station
order by l asc, c asc
limit 1;

或使用union：

(select city, length(city) 
from station 
order by length(city) asc , city asc limit 1) 
union
(select city,length(city) 
from station 
order by length(city) desc, city asc limit 1)

Answer 2

SELECT CITY,LENGTH(CITY) FROM STATION ORDER BY LENGTH(CITY) DESC,CITY ASC LIMIT 1;

SELECT CITY,LENGTH(CITY) FROM STATION ORDER BY LENGTH(CITY) ASC,CITY ASC LIMIT 1; 

/* FOR MYSQL */

Answer 3

使用以下查询：

(Select * from (Select city, length(city) from station order by length(city), city) where rownum = 1) Union All
(Select * from (Select city, length(city) from station order by length(city) desc, city) where rownum = 1);

Answer 4

我在这个问题上的成功是：-

ais-refinement-list

Answer 5

实际上，您的代码似乎正确。我认为唯一的问题可能是将工作地点设置为“ MYSQL”。如果您在“ MS SQL Server”上运行代码，它将给您一些“内置函数”的问题（例如，在mysql上，它写为lengt（），在ms sql服务器上，则是写为len（））（或“限制1”等）

我尝试过的另一种解决方案是（在MS SQL Server上）;

用于查找最长的字符城市（按字母顺序排在首位）；

Select TOP 1 city, LEN(CITY)
From station 
Where  
len(city) = (select max(len(city)) from station ) 
Order By city asc ;

用于查找最短的字符城市（按字母顺序排在首位）；

Select TOP 1 city, LEN(CITY)
From station
Where
len(city) = (select min(len(city)) from station) 
Order By city asc ;

Answer 6

SELECT city, CHAR_LENGTH(city) 
FROM station
ORDER BY CHAR_LENGTH(city), city
LIMIT 1;

SELECT city, CHAR_LENGTH(city) 
FROM station
ORDER BY CHAR_LENGTH(city) desc, city desc
LIMIT 1;

Answer 7

select t2.city , t2.t  
from  
(  
    select t1.city , t1.t , row_number() over (partition by t1.t order by t1.city) as ro 
    from
        ( select city , length(city)as t 
          from station 
        ) t1
    group by t1.city,t1.t
    having 
        t1.t = (select min(length(city)) from station )
                           or 
        t1.t = (select max(length(city)) from station)
) t2
where t2.ro = 1    ;

表t2将给出具有最小和最大字符串长度的所有记录以及行编号，现在基于行num过滤记录将获取您所需的输出

Answer 8

您也可以将此查询用作稍有不同的答案。我在MySQL的WHERE子句中使用了子查询

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

Answer 9

select * from  (select city,length(city) from station where  length(city) in(select min(length(city)) from station ) order by city asc) where rownum=1  ;

select * from (select city,length(city) from station where  length(city) in(select max(length(city)) from station )order by city asc) where rownum=1;

Answer 10

使用MySQL解决方案：

SELECT CITY, length(CITY) FROM STATION ORDER BY length( CITY), CITY ASC limit 1;

SELECT CITY, length(CITY) FROM STATION ORDER BY length(CITY) DESC limit 1;

Answer 11

关注MYSQL-

(SELECT  CITY , LENGTH(CITY) AS CITY_LENGTH
    FROM STATION
    ORDER BY CITY_LENGTH DESC, CITY ASC
    LIMIT 1)
UNION ALL
(SELECT  CITY , LENGTH(CITY) AS CITY_LENGTH
    FROM STATION
    ORDER BY CITY_LENGTH ASC, CITY ASC
    LIMIT 1)

关注评论以获取更好的解释- AS关键字-别名 ASC / DESC-升序/降序 LIMIT函数-限制查询输出联合函数-汇总结果

Answer 12

import Credenitals from 'your/path';

这对我有用，希望对您有所帮助。

Answer 13

对于甲骨文 ==> 我使用子查询解决了它并在子查询中使用了“MIN”和“MAX”函数，所以我跳过使用“ORDER BY”作为“MIN”，“MAX”函数在这里做同样的事情，只是使用“ORDER BY”外部查询按长度对最终输出进行排序。（希望有帮助！！）

SELECT CITY, LENGTH(CITY)
FROM STATION
WHERE CITY=(SELECT MIN(CITY)
            FROM STATION
            WHERE LENGTH(CITY)=(SELECT MIN(LENGTH(CITY))
                                FROM STATION)) OR
      CITY=(SELECT MIN(CITY)
            FROM STATION
            WHERE LENGTH(CITY)=(SELECT MAX(LENGTH(CITY))
                                FROM STATION))
ORDER BY LENGTH(CITY);

Answer 14

对于 Oracle - 使用密集排名：

SELECT CITY, LENGTH(CITY) FROM (
    SELECT CITY, LENGTH(CITY), 
        DENSE_RANK() OVER (ORDER BY LENGTH(CITY) ASC, CITY ASC) ROW1
    FROM STATION
)WHERE ROW1 = 1 
UNION
SELECT CITY, LENGTH(CITY) FROM (
    SELECT CITY, LENGTH(CITY), 
        DENSE_RANK() OVER (ORDER BY LENGTH(CITY) DESC, CITY ASC) ROW2
    FROM STATION
)WHERE ROW2 = 1;

Answer 15

(select city,length(city) from station 
order by  length(city) desc,city asc limit 1)
union
(select city,length(city) from station 
order by length(city) asc,city asc limit 1);

这在 MySQL 中有效。

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