Question

我试图使用mysql通过house_details表中的subincome字段显示progressive_total和cumulative_sum组。我在 link

中构建了我的架构

工作查询：

SELECT *,
       COALESCE(
                  (SELECT SUM(x.rupees)
                   FROM house_details x
                   WHERE MONTH(x.date) < t1.month), '-') AS progressive_total,

  (SELECT SUM(x.rupees)
   FROM house_details x
   WHERE MONTH(x.date) <= t1.month) AS cumulative_sum
FROM
  (SELECT MONTHNAME(t.date) AS `monthname`,
          MONTH(t.date) `month`,
          YEAR(t.date) AS YEAR,
          t.income,
          t.subincome,
          t.ssubincome,
          SUM(rupees) AS amount,
          GROUP_CONCAT(receipt_id) AS receipt_ids
   FROM house_details t
   WHERE YEAR(t.date) = YEAR(CURRENT_DATE())
   GROUP BY month(t.date),
            t.subincome
   ORDER BY t.date) t1

但这会在字段中显示无关的cumulative_sum。

我尝试在子查询中使用group，如下所示：

查询：

SELECT *,
       COALESCE(
                  (SELECT SUM(x.rupees)
                   FROM house_details x
                   WHERE MONTH(x.date) < t1.month
                   GROUP BY x.subincome), '-') AS progressive_total,

  (SELECT SUM(x.rupees)
   FROM house_details x
   WHERE MONTH(x.date) <= t1.month
   GROUP BY x.subincome) AS cumulative_sum
FROM
  (SELECT MONTHNAME(t.date) AS `monthname`,
          MONTH(t.date) `month`,
          YEAR(t.date) AS YEAR,
          t.income,
          t.subincome,
          t.ssubincome,
          SUM(rupees) AS amount,
          GROUP_CONCAT(receipt_id) AS receipt_ids
   FROM house_details t
   WHERE YEAR(t.date) = YEAR(CURRENT_DATE())
   GROUP BY month(t.date),
            t.subincome
   ORDER BY t.date) t1;

但它显示错误子查询返回多行。

Answer 1

在＆＃34; FROM＆＃34;的前一部分中写入的子查询必须只返回一行。显然，你的查询在这里转了多行。

此外，查询似乎有点复杂。您可以轻松获得这样的累进总和：

import java.io.BufferedReader;
import java.util.Scanner;

public class Main {

    public static void main(String[] args)
    {
        String myBeautifulScanf = new Scanner(System.in).nextLine();
        System.out.println( myBeautifulScanf );

    }
}

你可以按照这个想法分组：

每月分组

set @csum := 0;
select id, date, rupees ,(@csum := @csum + rupees) as proggressive_sum
from house_details
order by date;

分组年度和次级分组：

set @csum := 0;
select month,sum_rupees, (@csum := @csum + sum_rupees) as progressive_sum_monthly
from (
    select DATE_FORMAT(date,'%Y-%m') month, sum(rupees) sum_rupees
    from house_details
    GROUP BY DATE_FORMAT(date,'%Y-%m')
    ) gg
order by 1;

Answer 2

您可以使用以下查询来获得预期的结果集

SELECT *,
COALESCE(
    (SELECT SUM(pt.rupees) FROM (
        SELECT  MONTH(`date`) `month`,
        MAX(id) id,
        SUM(rupees) rupees
        FROM house_details 
        GROUP BY `month`,subincome
    ) pt 
    WHERE CASE WHEN pt.month = t1.month THEN pt.id < t1.id ELSE pt.month < t1.month END 
), 0) AS progressive_total,
(SELECT SUM(rupees) FROM(
        SELECT  MONTH(`date`) `month`,
        MAX(id) id,
        SUM(rupees) rupees
        FROM house_details 
        GROUP BY `month`,subincome
    ) cs 
    WHERE  CASE WHEN cs.month = t1.month THEN cs.id <= t1.id ELSE cs.month <= t1.month END 
) AS cumulative_sum 
FROM (
    SELECT MONTHNAME(t.date) AS `monthname`,
    MAX(id) id,
    MONTH(t.date) `month`,
    YEAR(t.date) AS `year`,
    GROUP_CONCAT(t.income) income,
    t.subincome,
    GROUP_CONCAT(t.ssubincome) ssubincome,
    SUM(rupees) AS amount,
    GROUP_CONCAT(receipt_id) AS receipt_ids 
    FROM house_details t 
    WHERE YEAR(t.date) = YEAR(CURRENT_DATE()) 
    GROUP BY `monthname`,`month`, t.subincome
    ORDER BY `month`
) t1

Demo

在子查询中使用group

2 个答案: