SQLite：运行 余额有期末余额
SQLite: Running balance with an ending balance
我的期末余额为 5000 美元。我需要创建一个 运行 余额，但调整第一行以显示期末余额，然后将其余部分相加，这样它看起来就像银行对帐单。这是 运行 余额的内容，但我如何调整第 1 行以不显示第一行的总和，而是显示期末余额。
with BalBefore as (
select *
from transactions
where ACCT_NAME = 'Real Solutions'
ORDER BY DATE DESC
)
select
DATE,
amount,
'$' || printf("%.2f", sum(AMOUNT) over (order by ROW_ID)) as Balance
from BalBefore;
这给了我
DATE AMOUNT BALANCE
9/6/2019 -31.00 $-31.00 <- I need this balance to be replaced with 00 and have the rest
9/4/2019 15.00 $-16.00 sum as normal.
9/4/2019 15.00 $-1.00
9/3/2019 -16.00 $-17.00
我已经阅读了很多其他问题，但找不到一个我能理解的问题，所以我想 post 一个更简单的问题。
以下内容并不简短，但使用 WITH statement and CTEs，希望逻辑清晰。定义了多个 CTE，它们相互引用以使整个查询更具可读性。总的来说，目标只是添加一个开始余额记录，该记录可能是：
/*
DROP TABLE IF EXISTS data;
CREATE temp TABLE data (
id INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT,
date DATETIME NOT NULL,
amount NUMERIC NOT NULL
);
INSERT INTO data
(date, amount)
VALUES
('2019-09-03', -16.00),
('2019-09-04', 15.00),
('2019-09-04', 15.00),
('2019-09-06', -31.00)
;
*/
WITH
initial_filter AS (
SELECT id, date, amount
FROM data
--WHERE ACCT_NAME = 'Real Solutions'
),
prepared AS (
SELECT *
FROM initial_filter
UNION ALL
SELECT
9223372036854775807 as id, --largest signed integer
(SELECT MAX(date) FROM initial_filter) AS FinalDate,
-(5000.00) --ending balance (negated for summing algorithm)
),
running AS (
SELECT
id,
date,
amount,
SUM(-amount) OVER
(ORDER BY date DESC, id DESC
RANGE UNBOUNDED PRECEDING
EXCLUDE CURRENT ROW) AS balance
FROM prepared
ORDER BY date DESC, id DESC
)
SELECT *
FROM running
WHERE id != 9223372036854775807
ORDER BY date DESC, id DESC;
这会产生以下结果
id date amount balance
4 2019-09-06 -31.00 5000
3 2019-09-04 15.00 5031
2 2019-09-04 15.00 5016
1 2019-09-03 -16.00 5001
更新：第一个查询没有产生正确的余额。更新期初余额行和窗口函数（即 OVER 子句）以准确计算正确金额的总和。
注意：每一行的余额完全由前几行确定，而不是当前行的金额，因为这是从期末余额反向计算的, 不是从上一行余额向前。
我的期末余额为 5000 美元。我需要创建一个 运行 余额，但调整第一行以显示期末余额，然后将其余部分相加，这样它看起来就像银行对帐单。这是 运行 余额的内容，但我如何调整第 1 行以不显示第一行的总和，而是显示期末余额。
with BalBefore as (
select *
from transactions
where ACCT_NAME = 'Real Solutions'
ORDER BY DATE DESC
)
select
DATE,
amount,
'$' || printf("%.2f", sum(AMOUNT) over (order by ROW_ID)) as Balance
from BalBefore;
这给了我
DATE AMOUNT BALANCE
9/6/2019 -31.00 $-31.00 <- I need this balance to be replaced with 00 and have the rest
9/4/2019 15.00 $-16.00 sum as normal.
9/4/2019 15.00 $-1.00
9/3/2019 -16.00 $-17.00
我已经阅读了很多其他问题，但找不到一个我能理解的问题，所以我想 post 一个更简单的问题。
以下内容并不简短，但使用 WITH statement and CTEs，希望逻辑清晰。定义了多个 CTE，它们相互引用以使整个查询更具可读性。总的来说，目标只是添加一个开始余额记录，该记录可能是：
/*
DROP TABLE IF EXISTS data;
CREATE temp TABLE data (
id INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT,
date DATETIME NOT NULL,
amount NUMERIC NOT NULL
);
INSERT INTO data
(date, amount)
VALUES
('2019-09-03', -16.00),
('2019-09-04', 15.00),
('2019-09-04', 15.00),
('2019-09-06', -31.00)
;
*/
WITH
initial_filter AS (
SELECT id, date, amount
FROM data
--WHERE ACCT_NAME = 'Real Solutions'
),
prepared AS (
SELECT *
FROM initial_filter
UNION ALL
SELECT
9223372036854775807 as id, --largest signed integer
(SELECT MAX(date) FROM initial_filter) AS FinalDate,
-(5000.00) --ending balance (negated for summing algorithm)
),
running AS (
SELECT
id,
date,
amount,
SUM(-amount) OVER
(ORDER BY date DESC, id DESC
RANGE UNBOUNDED PRECEDING
EXCLUDE CURRENT ROW) AS balance
FROM prepared
ORDER BY date DESC, id DESC
)
SELECT *
FROM running
WHERE id != 9223372036854775807
ORDER BY date DESC, id DESC;
这会产生以下结果
id date amount balance
4 2019-09-06 -31.00 5000
3 2019-09-04 15.00 5031
2 2019-09-04 15.00 5016
1 2019-09-03 -16.00 5001
更新：第一个查询没有产生正确的余额。更新期初余额行和窗口函数（即 OVER 子句）以准确计算正确金额的总和。
注意：每一行的余额完全由前几行确定，而不是当前行的金额，因为这是从期末余额反向计算的, 不是从上一行余额向前。