给定一年和日历周,我如何获得该周的星期二作为日期?
假设您有 year
和 cw
(日历周)作为变量(例如来自 SELECT 语句)你可以得到如下日期:
DATE_SUB(DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK),间隔工作日(DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK)) -1天),
短语DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK)
重复;不想存储变量.SQL 语句在 MySQL 上非常适合我.
更新: 只是为了澄清:WEEKDAY(DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK))
将产生一周的第一天.从中减去一个数字(-1 代表星期二;-2 代表星期三,依此类推,将为您选择一周中的特定日期).请参阅此处.>
Given a year and calendar week, how can I get the tuesday of that week as a date?
Given you have year
and cw
(calender week) as variables (e.g. from a SELECT statement) you can get the DATE as following:
DATE_SUB(
DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK),
INTERVAL WEEKDAY(
DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK)
) -1 DAY),
The phrase DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK)
is duplicated; did not want to store a variable. The SQL-Statement worked nicely for me on MySQL.
UPDATE: Just for clarification: WEEKDAY(DATE_ADD(MAKEDATE(year, 1), INTERVAL cw WEEK))
will yield the first day of the week. Substracting a number from it (-1 for Tuesday; -2 for Wednesday and so forth will select a specific day in the week for you).
See here.
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