bash Shell REGEX 表示日期
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Shell REGEX for date
提问by swag antiswag
I am trying to fix the users input if they input an incorrect date format that is not (YYYY-MM-DD) but I cant figure it out. Here is what I have:
如果用户输入的日期格式不正确（YYYY-MM-DD），我正在尝试修复用户输入，但我无法弄清楚。这是我所拥有的：
while [ "$startDate" != "^[0-9]{4}-[0-9]{2}-[0-9]{2}$" ]
do
echo "Please retype the start date (YYYY-MM-DD):"
read startDate
done
回答by fedorqui 'SO stop harming'
Instead of !=, you have to use ! $var =~ regexto perform regex comparisons:
取而代之的是!=，您必须使用! $var =~ regex来执行正则表达式比较：
[[ $date =~ ^[0-9]{4}-[0-9]{2}-[0-9]{2}$ ]]
^^
So that your script can be like this:
这样你的脚本就可以是这样的：
date=""
while [[ ! $date =~ ^[0-9]{4}-[0-9]{2}-[0-9]{2}$ ]]; do
echo "enter date (YYYY-MM-DD)"
read $date
done
回答by Wei Shen
Why not convert the incorrect format to desired one using date program:
为什么不使用日期程序将不正确的格式转换为所需的格式：
$ date -d "06/38/1992" +"%Y-%m-%d"
1992-06-28
You can also check for conversion failure by checking return value.
您还可以通过检查返回值来检查转换失败。
$ date -d "06/38/1992" +"%Y-%m-%d"
date: invalid date ‘06/38/1992'
$ [ -z $? ] || echo "Error, failed to parse date"
