Question

这是针对数据库（https://www.hackerrank.com/challenges/weather-observation-station-5）的HackerRank Weather Observation 5问题。我该如何解决这个问题？

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

这是我到目前为止所拥有的

SELECT CITY, MAX LENGTH(CITY) FROM STATION;

但它显然不起作用。

Answer 1

这是一个带窗函数的解决方案：

select city, length(city)
from
(
  select 
    city, 
    row_number() over (order by length(city), city) as shortest_is_one,
    row_number() over (order by length(city) desc, city) as longest_is_one
  from station
)
where shortest_is_one = 1 or longest_is_one = 1;

替代方案是使用UNION的{{1}}查询：

FETCH FIRST 1 ROW ONLY

这又是另一个查询，再次使用( select city, length(city) from station order by length(city), city fetch first 1 row only ) union ( select city, length(city) from station order by length(city) desc, city fetch first 1 row only );：

UNION

Answer 2

在单表扫描中：

SELECT MIN( city ) KEEP ( DENSE_RANK FIRST ORDER BY LENGTH( city ) ) AS shortest_city,
       MIN( city ) KEEP ( DENSE_RANK LAST  ORDER BY LENGTH( city ) ) AS longest_city,
       LENGTH( MIN( city ) KEEP ( DENSE_RANK FIRST ORDER BY LENGTH( city ) ) )
         AS shortest_length,
       LENGTH( MIN( city ) KEEP ( DENSE_RANK LAST  ORDER BY LENGTH( city ) ) )
         AS longest_length
FROM   station;

<强>解释

使用... KEEP ( DENSE_RANK [FIRST|LAST] ORDER BY ...)时，您可以从右向左阅读以评估其效果。

所以：

MIN( city ) KEEP ( DENSE_RANK FIRST ORDER BY LENGTH( city ) )

威尔：

ORDER BY子句按城市长度排序结果;然后
KEEP ( DENSE_RANK FIRST子句仅保留在顺序中排名第一的结果（并且如果有多个结果在顺序中排在第一位，那么它将保留所有首先联合的结果）;最后，
MIN( city )将从保留的结果（即只有长度最短的结果）中返回city的最小值（即字母顺序的名字）。

Answer 3

这是我的解决方案。我不确定性能，但它对我来说效果很好，而且不难理解。基本上我们只是选择城市和城市的长度并将其存储在var'len_city'中，诀窍是我们将它们列为升序，但将列表限制为一个。然后我们用相同的查询结合该答案，这次，通过降序排序，但也限于一个。这样我们得到第一个最短和最后一个。我对更简单的实现持开放态度，但这不是火箭科学，其他可能更难理解但更有效。我对此很新，但不是编程，所以，希望这有帮助！

(SELECT city, CHAR_LENGTH(city) AS len_city FROM station ORDER BY len_city ASC LIMIT 1)
UNION ALL
(SELECT city, CHAR_LENGTH(city) AS len_city FROM station ORDER BY len_city DESC LIMIT 1)

Answer 4

(select min(city) count(min(city)) from station order by city )
union all
(select max(city) count(max(city)) from station order by city desc);

Answer 5

(SELECT city, CHAR_LENGTH(city) as val1 FROM station order by  CHAR_LENGTH(city) ASC limit 0,1) 
UNION ALL 
(SELECT city, CHAR_LENGTH(city) as val2 FROM station order by CHAR_LENGTH(city) DESC limit 0,1)

Answer 6

select CITY
  , length(CITY)
from station
order by length(CITY), city
limit 1; 

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

查找字符串的最大长度+ Oracle SQL中的字符串

6 个答案: