Question

我在Hackerrank上做了这个挑战：

https://www.hackerrank.com/challenges/weather-observation-station-5

我是SQL的初学者，我尝试查询列中具有最大值的所有行，我只能通过MAX()获取最大值。所以我试试这个：

SELECT CITY, LENGTH(CITY) AS citylength
FROM STATION
WHERE LENGTH(CITY) = (SELECT MIN(CITY) FROM STATION)

我收到错误。

我已经在google上查询了有关子查询的内容，但我并不习惯知道它究竟是如何工作的，所以我需要你的帮助。谢谢。

总而言之，我需要一个查询，可以获取通过MAX()子句获得最大值的表上的行。

Answer 1

要求您找到两种不同的结果：

具有最大长度的城市（如果是平局，则为字母表中的第一个）
长度最小的城市（如果是领带，则为字母表中的第一个）

这意味着您将两个不同的查询与UNION ALL粘合在一起。

(
  select concat(city, ' ', length(city))
  from station 
  order by length(city), city limit 1
)
union all
(
  select concat(city, ' ', length(city))
  from station 
  order by length(city) desc, city limit 1
);

正如Strawberry指出的那样：你需要括号来放置两个ORDER BY子句，每个查询部分一个。（否则，您只能在整个查询的末尾放置一个ORDER BY子句。）

在您的查询中，您正在比较LENGTH(CITY)，即一个包含名称长度的整数和MIN(CITY)，即城市名称本身，当然无法正常工作。您必须与MIN(LENGTH(CITY))进行比较。然后执行相同的最大值，然后使用UNION ALL。但这并不能解决LIMIT查询所带来的关系问题。

Answer 2

这没有子查询

SELECT CITY, LENGTH(CITY)
FROM STATION
ORDER BY 2,1
LIMIT 1

Answer 3

这有效：

SELECT CITY,LENGTH(CITY) 
FROM STATION  
WHERE LENGTH(CITY) = (SELECT MIN(LENGTH(CITY)) M FROM STATION);

Answer 4

这可以正常工作：

select city, length(city) 
from station 
where length(city) = (select min(length(city)) as m from station)
order by city ASC
limit 1;

Answer 5

ArrayList<Beacon> arrayList = new ArrayList<Beacon>();
String beaconName;

public void addToBeaconInfo(Beacon beacon) {

    boolean beaconAlreadyInList = false;
    //arrayList.add(beacon);
    beaconName = "" + beacon.getId2() + "" + beacon.getId3();
    //Log.i("BeaconInfo","addBeacon_beaconName: "+beaconName + " distance" + beacon.getDistance());

    int i;
    for(i = 0; i < arrayList.size(); i++){
        Beacon tempBeacon = arrayList.get(i);

        if(tempBeacon.getId2().equals(beacon.getId2())){
            beaconAlreadyInList = true;
            arrayList.remove(beacon);

            Log.i("BeaconInfo", "addBeacon_tempBeacon: " + beacon.getId3() + " distance: " + beacon.getDistance());
            break;
        }
    }
    if(beaconAlreadyInList == false){
        arrayList.add(beacon);
        Log.i("arrayList","" + beacon.getDistance() + " " + beacon.getBluetoothName());
        double maxDistance = beacon.getDistance();
        //double a = Collections.max(maxDistance);
        //Log.i("maxDistance","" + Math.max());
    }
}

Answer 6

select TOP 1 concat(city, ' ' , len(city)) from station where len(city) = (select min(len(city)) from station) order by city asc;

select TOP 1 concat(city, ' ' , len(city)) from station where len(city) = (select max(len(city)) from station) order by city desc;

Answer 7

SELECT * FROM
    (SELECT City, LENGTH(City) FROM STATION 
    ORDER BY LENGTH(City),city ASC) 
WHERE ROWNUM <= 1
UNION ALL
SELECT * FROM 
    (SELECT City, LENGTH(City) FROM STATION 
    ORDER BY LENGTH(City) DESC) 
WHERE ROWNUM <= 1;

如何选择具有列的最大值的行

7 个答案: