Question

SELECT DATE_PART('DAY',"repairClose"-"createdOn")::INTEGER as "noOfDays"
FROM "mstRecord"
WHERE "repairClose" is not null

我尝试了上面的查询，它返回包括星期日的结果。我希望结果排除星期日。

Answer 1

尝试使用extract(dow from "column") <> 0

这是一个例子：

t=# with dates as (select generate_series(now(),now()+'3 weeks'::interval,'1 day'::interval) "repairClose", now() "createdOn")
select "repairClose"-"createdOn","repairClose"::date, "createdOn"::date 
from dates 
where extract(dow from "repairClose") <> 0;
 ?column? | repairClose | createdOn  
----------+-------------+------------
 00:00:00 | 2017-07-31  | 2017-07-31
 1 day    | 2017-08-01  | 2017-07-31
 2 days   | 2017-08-02  | 2017-07-31
 3 days   | 2017-08-03  | 2017-07-31
 4 days   | 2017-08-04  | 2017-07-31
 5 days   | 2017-08-05  | 2017-07-31
 7 days   | 2017-08-07  | 2017-07-31
 8 days   | 2017-08-08  | 2017-07-31
 9 days   | 2017-08-09  | 2017-07-31
 10 days  | 2017-08-10  | 2017-07-31
 11 days  | 2017-08-11  | 2017-07-31
 12 days  | 2017-08-12  | 2017-07-31
 14 days  | 2017-08-14  | 2017-07-31
 15 days  | 2017-08-15  | 2017-07-31
 16 days  | 2017-08-16  | 2017-07-31
 17 days  | 2017-08-17  | 2017-07-31
 18 days  | 2017-08-18  | 2017-07-31
 19 days  | 2017-08-19  | 2017-07-31
 21 days  | 2017-08-21  | 2017-07-31
(19 rows)

Answer 2

import java.util.Scanner;

public class DecreasingOrNot {

    public static void main(String[] args){
        int number1, number2;
        boolean decrease = true;
        Scanner input = new Scanner(System.in);
        System.out.println("Enter a sequence of numbers: ");
        number2 = input.nextInt();
        while (decrease && input.hasNext()){
            number1 = number2;
            number2 = input.nextInt();
            if (number1 < number2){
                decrease = false;
            }
        }
        if (decrease){
            System.out.println("Yes");
        }
        else{
            System.out.println("No");
        }
    }
}

Answer 3

选择（SELECT count（*）AS count_days_no_weekend FROM generate_series（“createdOn”:: date，“repairClose”:: date，'1 day'）d（the_day）提取物（'ISODOW' FROM the_day）＆lt; 7）作为“no_of_days” 来自“mstRecord”，其中“repairClose”不为null 按“no_of_days”排序

在postgres中，将“isodow”选项添加到EXTRACT（），其中Sunday = 7，Monday = 1，Tuesday = 2，依此类推。

计算工作日（不包括星期日）

3 个答案: