我在雪花中有一个视图，它给了我以下信息：
如果你愿意的话，我想从中生成一种摊销计划。因此，如果我的贷款日期为 2022 年 1 月 1 日，到期日为 2022 年 3 月 9 日，付款频率为每两周一次，每次付款 50 美元，我希望看到如下输出：
| 贷款ID | 付款日期 | 支付金额 | 付款频率 |
|---|---|---|---|
| abc123 | 2022 年 1 月 15 日 | 50.00 美元 | 双周刊 |
| abc123 | 2022 年 1 月 29 日 | 50.00 美元 | 双周刊 |
| abc123 | 2/12/2022 | 50.00 美元 | 双周刊 |
| abc123 | 2022 年 2 月 26 日 | 50.00 美元 | 双周刊 |
| abc123 | 2022 年 3 月 9 日 | 50.00 美元 | 双周刊 |
我假设我需要某种循环，而付款日期<到期日期和总和（付款金额）<贷款金额，但我不知道如何正确设置它以查看数千笔贷款。你们能提供的任何帮助都会令人难以置信，我非常感激！
您可以通过编写Recursive CTE来获得它，只需记住默认限制为 100 次迭代，如果您需要更多循环，请检查此MAX_RECURSIONS参数。
这只是一个代码示例，您应该对其进行扩展以包含一些极端的数据保护；
样本数据：
CREATE OR REPLACE TABLE LoanTable (
LoanID STRING,
Loan_date DATE,
Loan_amount NUMERIC(12,2),
Maturity_date DATE,
Payment_frequency STRING,
Payment_amount NUMERIC(12,2)
);
INSERT INTO LoanTable
VALUES ('abc123', '1/1/2022', 250, '3/9/2022', 'biweekly', 50);
询问：
WITH Recursive_CTE AS (
SELECT LoanID,
CASE Payment_frequency WHEN 'weekly' THEN DATEADD(WEEK, 1, Loan_date)
WHEN 'biweekly' THEN DATEADD(WEEK, 2, Loan_date)
WHEN 'semimonthly' THEN DATEADD(DAY, 15, Loan_date) -- I don't know how the semimonthly value is determined??
WHEN 'monthly' THEN DATEADD(MONTH, 1, Loan_date) END AS Payment_Date,
Payment_amount,
Loan_amount - Payment_amount AS Left_to_pay,
Payment_frequency,
Maturity_date
FROM LoanTable
UNION ALL
SELECT LoanID,
CASE Payment_frequency WHEN 'weekly' THEN DATEADD(WEEK, 1, Payment_Date)
WHEN 'biweekly' THEN DATEADD(WEEK, 2, Payment_Date)
WHEN 'semimonthly' THEN DATEADD(DAY, 15, Payment_Date) -- I don't know how the semimonthly value is determined??
WHEN 'monthly' THEN DATEADD(MONTH, 1, Payment_Date) END AS Payment_Date,
Payment_amount,
IFF(Left_to_pay - Payment_amount < 0, Left_to_pay, Left_to_pay - Payment_amount) AS Left_to_pay,
Payment_frequency,
Maturity_date
FROM Recursive_CTE
WHERE Left_to_pay > 0
)
SELECT LoanID, IFF(Payment_Date > Maturity_date, Maturity_date, Payment_Date) AS Payment_Date, Payment_amount, Left_to_pay, Payment_frequency
FROM Recursive_CTE
ORDER BY LoanID, Payment_Date;
表生成器是另一种方法。
WITH CTE_MY_DATE AS (
SELECT DATEADD(DAY, SEQ4(), '1900-01-01') AS MY_DATE
FROM TABLE(GENERATOR(ROWCOUNT => 180000)) -- Number of days after reference date in previous line
)
SELECT
date(MY_DATE) CALENDAR_DATE,
concat( decode(extract ('dayofweek_iso', date(MY_DATE)),1,'Monday',2, 'Tuesday',3, 'Wednesday',4, 'Thursday',5, 'Friday',6, 'Saturday',7, 'Sunday'),TO_CHAR(date(MY_DATE), ', MMMM DD, YYYY')
) FULL_DATE_DESC
,row_number() over (partition by 1 order by calendar_date ) modd
FROM
CTE_MY_DATE
where
CALENDAR_DATE
between '2022-01-02' and '2022-09-03'
qualify
mod(modd, 14) = 0
