繁体   English   中英

MySQL借方贷方余额

[英]mySQL debit credit balances

是否可以使用mySQL查询来获取这些字段的旧余额(我的支出总和),借方(我应该在这里将具有相同id的总支出放在当前日期),贷方(在这些查询中没有多少),新余额(与借记相同)。

现在,当一天结束时,新余额将添加到旧余额中,并且将为当前日期形成新余额。 我有我的代码,但它不计算我以前的余额。

表:rf_expenses

--------------------------------------
id_rf_expenses   rf_expense_desc
--------------------------------------
1                    salary
2                    bonus
3                    transportation

表:费用

-----------------------------------------------------------
id_expnses    id_rf_expenses     expense_amt   expense_date
------------------------------------------------------------
1                1                  100           current
2                1                  100           yesterday
3                2                   50           current
4                2                   50           current
5                3                  200           yesterday

输出:

-----------------------------------------------------------
EXPENSE      OLD BALANCE    DEBIT     CREDIT   NEW BALANCE
-----------------------------------------------------------
SALARY           100         100        0          100          
BONUS             0          100        0          100
TRANSPO          200           0        0            0

这将对当前日期进行每次查询。.当前日期之前的任何日期都将在旧余额中汇总

SELECT 
    r.rf_expense_desc, 
    COALESCE(SUM(expense_amt), 0) - COALESCE(q1.amt2, 0) OLD, 
    COALESCE(q1.amt2, 0) AS NEW
FROM 
    rf_expenses r
LEFT JOIN 
    expenses e ON r.id_RF_expense = e.id_rf_expense
LEFT JOIN 
    (SELECT 
         SUM(expense_amt) amt2, id_expense, 
         rf_expenses.id_rf_expense
     FROM 
         rf_expenses
     LEFT JOIN 
         expenses ON rf_expenses.id_RF_expense = expenses.id_rf_expense
     WHERE 
         expense_date = CURDATE( ) 
     GROUP BY 
         rf_expense_desc, expense_date) q1 ON r.id_rf_expense = q1.id_rf_expense
GROUP BY 
    rf_expense_desc 

嗨,我给出了两个查询:一个查询与日期类型列,另一个查询与字符串。

1查询(带有示例数据) SQLFIDDLE示例

    SELECT 
rf.rf_expense_desc as EXPENSE      
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date != 'current'
 ) AS 'OLD BALANCE'
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date = 'current'
 ) AS 'DEBIT'
,0 AS CREDIT
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date = 'current'
 ) AS 'NEW BALANCE'
FROM 
rf_expenses rf

2查询(数据类型为日期) SQLFIddle example2

SELECT 
rf.rf_expense_desc as EXPENSE      
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date != CURDATE()
 ) AS 'OLD BALANCE'
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date = CURDATE()
 ) AS 'DEBIT'
,0 AS CREDIT
,(SELECT 
  COALESCE(SUM(e.expense_amt),0)
  FROM expenses e
  WHERE e.id_rf_expenses= rf.id_rf_expenses
  AND expense_date = CURDATE()
 ) AS 'NEW BALANCE'
FROM 
rf_expenses rf

结果:

|        EXPENSE | OLD BALANCE | DEBIT | CREDIT | NEW BALANCE |
---------------------------------------------------------------
|         salary |         100 |   100 |      0 |         100 |
|          bonus |           0 |   100 |      0 |         100 |
| transportation |         200 |     0 |      0 |           0 |

mysql为什么要强制所有GROUP BY子句的SELECT仅具有GROUP BY子句中指定的列或由聚合函数包围的列的另一个示例。 Mysql将此错误称为“功能”。

我的意思是,看看两个GROUP BY的SELECT子句,内部查询和外部查询都一样。 您选择的列不在GROUP BY子句中,并且不被SUM,AVG,COUNT或其他聚合函数包围。 如果将行分组,并且不告诉mysql如何“减少”组中的行,它将简单地随机获取任何值。

例如,假设我们将这些数据存储在名为“ test”的表中:

PARENTID     ID     VALUE
-------------------------
1            1      2.05
1            2      3.50
1            3      1.58
2            1      4.65
2            2      0.65

我们查询:SELECT父标识,值FROM测试GROUP BY父标识,值

好吧,我们还没有说要对VALUE列进行处理,例如,对SUM进行处理,因此mysql将简单地获取组中的任何值,对于PARENTID = 1,我们可能会获得2.05、3.50或1.58中的任何一个...但不能合并很多。

所以这是正确的:

SELECT parentid, SUM(value) as total FROM test GROUP BY parentid, value;

现在,当mysql减少每个组时,我们得到正确的总和,我们的结果是:

PARENTID          VALUE
-------------------------
1                 7.13
2                 5.30

因此,在您的示例中,您可能要查询:

SELECT r.rf_expense_desc, expense_date,
    COALESCE( SUM( expense_amt ) , 0 ) - COALESCE( sum(q1.amt2), 0 ) OLD, COALESCE( sum(q1.amt2), 0 ) AS NEW
FROM rf_expenses r
LEFT JOIN expenses e ON r.id_RF_expense = e.id_rf_expense
LEFT JOIN (
    SELECT SUM( expense_amt ) amt2, count(id_expense) as counts
    FROM rf_expenses
    LEFT JOIN expenses ON rf_expenses.id_RF_expense = expenses.id_rf_expense
    WHERE expense_date = CURDATE( ) 
    GROUP BY rf_expense_desc, expense_date
) q1 ON r.id_rf_expense = q1.id_rf_expense
GROUP BY rf_expense_desc 

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM