在 bash 脚本中获取星期几
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Get Day Of Week in bash script
提问by bbholzbb
I want to have the day of week in the variable DOW.
我想在变量中包含星期几DOW。
So I use the following bash-script:
所以我使用以下 bash 脚本：
DOM=$(date +%d)
DOW=($($DOM % 7) ) | sed 's/^0*//'
Unfortunately there I get this error: bash: 09: command not found. The expected result is 2 ( 9 % 7 = 2) in the variable $DOW.
不幸的是，我收到此错误：bash: 09: command not found。变量中的预期结果是 2 ( 9 % 7 = 2) $DOW。
How can I get this working?
It works for the days 1-8 but because of the C-hex, there is no number over 8 available and the following message appears: bash: 09: value too great for base (error token is "09").
我怎样才能让它工作？它适用于 1-8 天，但由于 C-hex，没有超过 8 的可用数字，并显示以下消息：bash: 09: value too great for base (error token is "09")。
回答by hek2mgl
回答by glenn Hymanman
Using a different %-specifier is the real answer to your question. The way to prevent bash from choking on invalid octal numbers is to tell it that you actually have a base-10 number:
使用不同的 % 说明符是您问题的真正答案。防止 bash 因无效八进制数而窒息的方法是告诉它您实际上有一个以 10 为基数的数字：
$ DOM=09
$ echo $(( DOM % 7 ))
bash: 09: value too great for base (error token is "09")
$ echo $(( 10#$DOM % 7 ))
2
回答by Ronald Hofmann
This works fine here
这在这里工作正常
#!/bin/sh
DOW=$(date +"%a")
echo $DOW
回答by LordWilmore
I've always found that using 'let' is the most simple solution for BASH. So that would be:
我一直发现使用“让”是 BASH 最简单的解决方案。所以那将是：
let "DOW = DOM % 7"
回答by devnull
You can use the -flag:
您可以使用-标志：
DOM=$(date +%-d)
^
which would prevent the day from being padded with 0.
这将防止一天被填充0。
From man date:
来自man date：
- (hyphen) do not pad the field
Observe the difference:
观察差异：
$ DOM=$(date +%d)
$ echo $((DOM % 7))
bash: 09: value too great for base (error token is "09")
$ DOM=$(date +%-d)
$ echo $((DOM % 7))
2
