选择最小值和最大值

时间:2019-11-01 19:36:56

标签: mysql sql max min

我有一个名为“客户”的表:

id | name    | age

1  | john    | 35        
2  | paul    | 22        
3  | ana     | 26   
4  | mark    | 19   
5  | jack    | 29   

我想选择姓名和最大年龄,姓名和最小年龄...类似

john 35 mark 19

有可能吗?

6 个答案:

答案 0 :(得分:1)

以下查询将根据您的要求在一行中提供最大和最小。如果最小/最大有多个匹配项,您将获得多行。取决于所使用的SQL引擎,限制为一行的语法是不同的。

SELECT cMax.Name, cMax.Age, cMin.Name, cMin.Age
FROM customers cMin
JOIN customers cMax ON
    cMax.Age = (SELECT MAX(Age) FROM customers)
WHERE cMin.Age = (SELECT MIN(Age) FROM customers)

有不同类型的联接(例如INNER,OUTER,CROSS);但是,对于您的问题,使用哪个都没关系。

答案 1 :(得分:0)

尝试

    select name, age from customers where age=(select max(age) from customers) union
 select name, age from customers where age=(select min(age) from customers) 

答案 2 :(得分:0)

是的,你可以做到

select name, age from customers
where age in (select max(age) 
from customers union select min (age)from customers)

答案 3 :(得分:0)

如果您希望它们位于同一行:

private PostCallback callback;
private Exception exception;

public setPostCallback(PostCallback callback){
    this.callback = callback;
}

@Override
protected JSONObject doInBackground(String... args){
     // keep everything as before but when an Exception occurs,
     // assign it to *exception* in the catch block
} 

@Override
public void onPostExecute(JSONObject jsonObject){
    // Note: this will be executed on the main thread
    if(exception == null){
        callback.onSuccess(jsonObject);
    } else {
        callback.onError(exception);
    }
}

select cmin.*, cmax.* from (select name, age as minage from customers order by age asc fetch first 1 row only ) cmin cross join (select name, age as maxage from customers order by age desc fetch first 1 row only ) cmax; 是仅返回结果集第一行的标准语法。某些数据库具有定制语法,例如fetch first 1 row onlylimit

答案 4 :(得分:0)

尝试使用此查询显示MAX年龄:-     select * from customers where age=(select max(age) from customers);

要显示MIN年龄,请使用以下查询:-     select * from customers where age=(select min(age) from customers);

答案 5 :(得分:-1)

您可以使用cross join,它将两个查询输出并排放置。建立Rodrigo的查询:

df %>%
  group_by(as.Date(Datetime)) %>%
  mutate(LightIntensity = Irradiance/max(Irradiance)) 

它可能不是性能最高的,但形状正确。当表的行不完全为1时,请小心交叉联接,因为它会创建要联接的表中各行的笛卡尔积。

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