HackerRank天气观测站5

Posted 2021-04-28

我想知道为什么我的代码不起作用。这个问题在此之前已经被问到：qazxsw poi

和解决方案：Query the two cities in STATION with the shortest and longest CITY names,

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。谢谢您的帮助！

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城市。

输入格式

STATION表描述如下：

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

其中LAT_N是北纬，LONG_W是西经。

样本输入

假设CITY只有四个条目：DEF，ABC，PQRS和WXY

样本输出

Station.jpg

说明

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。

注意您可以编写两个单独的查询以获得所需的输出。它不一定是一个查询。

我的答案：

/最短字符长度按字母顺序排序/

ABC 3
PQRS 4

/最长字符长度按字母顺序排序/

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

答案

您在github上的解决方案如下所示：

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

你有一个问题 - 没有像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; 这样的命令。根据数据库系统，它可以是fetch first row only，top或limit - 请在这里阅读更多 - rownum

因此，根据系统的不同，答案也会有所不同。

神谕

https://www.w3schools.com/sql/sql_top.asp

SQL Server

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;

mysql的

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;

或使用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; ：

union

另一答案

使用以下查询：

(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)

另一答案

我对这个问题的看法是： -

(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);

另一答案

实际上你的代码似乎是对的。我认为唯一的问题可能是设置工作地点而不是“MYSQL”。如果您在“MS SQL Server”上运行代码，它将为您提供一些“内置函数”问题（比如在mysql上编写lengt（），但在ms sql server上写入len（））（或“限制1”）等等。）

我试过的其他解决方案之一是（在MS SQL Server上）;

寻找最长的角色城市（按字母顺序排列）;

SELECT CITY,LENGTH(CITY)

FROM STATION

ORDER BY LENGTH(CITY),CITY

LIMIT 1 OFFSET 0;



SELECT CITY,LENGTH(CITY)

FROM STATION

ORDER BY LENGTH(CITY) DESC,CITY

LIMIT 1 OFFSET 0;

寻找最短的人物城市（按字母顺序排列）;

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

另一答案

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

另一答案

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;

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过滤记录将获取所需的输出