如何使用 Postgresql 对 table 的所有列值求和并在新行中显示总数

Question

我正在使用正常的 select 查询来显示所有行

SELECT 
    type,
    debit,
    credit,
    (debit-credit) as balance
from bank_cash_registers

显示如下图

现在我需要在 Postgresql 查询的帮助下将此总计显示为附加行，如下图所示。我怎样才能做到这一点？

还有是否有任何选项可以根据类型单独计算总计，如下所示。

Answer 1

SELECT type
    ,debit
    ,credit
    ,(debit - credit) as balance
FROM bank_cash_register

UNION ALL

SELECT 'Total: '
    ,sum(debit)
    ,sum(credit)
    ,sum((debit - credit))
FROM bank_cash_register

根据类型列对合计进行分组

SELECT *
FROM (
    SELECT type
          ,debit
          ,credit
          ,(debit - credit) as balance
    FROM bank_cash_register

    UNION ALL

    SELECT type || '_total'
          ,sum(debit)
          ,sum(credit)
          ,sum((debit - credit))
    FROM bank_cash_register
    GROUP BY 1
    ) t
ORDER BY split_part(type, '_', 1)

通过split_part()

Demo

Answer 2

另一种方法是使用 grouping sets。它的优点是易于扩展。此外，我认为它是专门为此目的创建的。

这应该比 UNION 解决方案更有效，因为数据只传递一次。

下面查询returns你想要什么：

SELECT COALESCE(type, 'Total: '), SUM(debit), SUM(credit), SUM(debit - credit) AS balance
FROM bank_cash_registers
GROUP BY GROUPING SETS ((type, debit, credit), ());

以下查询将具有相同类型的值组合在一起（请注意，唯一改变的是 GROUPING SETS 子句）：

SELECT COALESCE(type, 'Total: '), SUM(debit), SUM(credit), SUM(debit - credit) AS balance
FROM bank_cash_registers
GROUP BY GROUPING SETS ((type), ());

结果：

bank    0    1500    -1500
cash    0     700     -700
Total:  0    2200    -2200

您更新后的问题可以通过以下方式解决：

SELECT
  CASE WHEN GROUPING(debit) > 0 THEN 'Total: ' ELSE type END AS type,
  SUM(debit), SUM(credit), SUM(debit - credit) AS balance
FROM bank_cash_registers
GROUP BY GROUPING SETS ((type, debit, credit), (type));

您甚至可以用

加总

(...) GROUPING SETS ((type, debit, credit), (type), ());

如何使用 Postgresql 对 table 的所有列值求和并在新行中显示总数

How to sum all columns value of a table and display the total in new row by using Postgresql

sql

postgresql

postgresql-9.3

postgresql-9.4