我是Oracle的新手。我一直在研究如何用to_Date减去两天之间的例子。
但是，每次我使用to_date logic 运行它
SELECT (to_date((SELECT PAYMENT_DATE FROM INVOICES WHERE INVOICE_ID =3) - to_date((SELECT INVOICE_DUE_DATE FROM INVOICES WHERE INVOICE_ID= 3)))) FROM DUAL; 我最终收到此错误：ORA-01847：月中的某一天必须介于1和最后一天之间
付款日期为：15-SEP-17发票到期日为：24-JUN-17
有没有一种简单的方法来计算两列之间的天数差异？ 提前致谢。
答案 0 :(得分：0)
您不需要使用SELECT *,
CASE WHEN selection = 'All Assets'
THEN 'Other Assets'
WHEN (isassetcode = 1 AND selection <> 'All Assets')
THEN selection
END AS Level5,
COUNT(*) cnt
FROM table --or sub-query
GROUP BY list all columns in the select that corresponds the *,
CASE WHEN selection = 'All Assets'
THEN 'Other Assets'
WHEN (isassetcode = 1 AND selection <> 'All Assets')
THEN selection
END
进行转换，而无需像to_date这样的格式。即省略'dd.mm.rrrr'条款。减去日期会产生一个数字（假设PAYMENT_DATE，INVOICE_DUE_DATE列都在to_date类型中）
date其中v_day是一个数字。顺便说一下，以这种方式使用会更好：
SELECT (SELECT PAYMENT_DATE FROM INVOICES WHERE INVOICE_ID =3) - (SELECT INVOICE_DUE_DATE FROM INVOICES WHERE INVOICE_ID= 3) INTO v_day FROM DUAL;