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我有一个自定义列表,我试图限制有效星期几和时间的数据输入。我当前的列验证适用于星期一、星期三或星期五的星期几。它看起来像这样:

=CHOOSE(WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE)

我正在尝试找出语法来添加它也必须在那些日子的上午 8 点到下午 12:00 之间。

任何帮助将不胜感激。

4

1 回答 1

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您将使用 AND 语句来包含第二个条件

=AND(CHOOSE(WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE),
  AND(
    [Requested date for approval]-INT([Requested date for approval])*24 >= 8,
    [Requested date for approval]-INT([Requested date for approval])*24 <= 24
  )
)

我承认,我从未听说过 CHOOSE 功能,但时间计算是基于 Microsoft 的信息

转换时间 要将小时从标准时间格式转换为十进制数,请使用 INT 函数。

Column1       公式                           说明(可能的结果)
10:35 AM =([Column1]-INT([Column1]))*24 自 12:00 AM (10.583333) 以来的小时数
12:15 PM =([Column1]-INT([Column1]))*24 自 12:00 AM (12.25) 以来的小时数

编辑

要计算星期几,可以使用 TEXT 函数返回星期几(即星期一)

=TEXT(WEEKDAY([ColumnName]), "dddd")

它不会很漂亮,但是您可以使用一系列 AND 逻辑运算符

=AND(
  TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Monday",
  AND(
    TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Wednesday",
    AND(
      TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Friday",
      AND(
        [Requested date for approval]-INT([Requested date for approval])*24 >= 8,
        [Requested date for approval]-INT([Requested date for approval])*24 <= 24
      )
    )
  )
)

发布工作解决方案

=IF(
  AND(
    CHOOSE(
      WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE
    ),
    ([Requested date for approval]-INT([Requested date for approval]))*24>=8,
    ([Requested date for approval]-INT([Requested date for approval]))*24<=12
  ),
  TRUE
)
于 2015-04-27T21:03:09.103 回答