Bash 验证日期
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Bash validate date
提问by Cbas
I'm writing a shell script and am confused as to why my date validation code is not working. I tried the following solutions to similar questions I found, but is_valid is always set to 1:
我正在编写一个 shell 脚本，对为什么我的日期验证代码不起作用感到困惑。我针对我发现的类似问题尝试了以下解决方案，但 is_valid 始终设置为 1：
date "+%m/%d/%Y" -d "" 2>1 > /dev/null
//or..
date -d "2012-02-29" > /dev/null 2>&1
is_valid=$?
#always set to 1, even when given a valid date
How do I correctly validate the date format? The date should only be valid if in the format MM/DD/YYYY
如何正确验证日期格式？日期只有在格式为 MM/DD/YYYY 时才有效
I also tried this solution: Linux Bash - Date Formatbut it always rejected the date as well.
我也尝试过这个解决方案：Linux Bash - Date Format但它也总是拒绝日期。
回答by chepner
The BSD datethat ships with Mac OS X doesn't support the -doption (or rather, it uses -dfor something entirely different). Either install GNU date, or use the following to validate your input string:
dateMac OS X 附带的 BSD不支持该-d选项（或者更确切地说，它-d用于完全不同的东西）。要么安装 GNU date，要么使用以下内容来验证您的输入字符串：
date -f "%Y-%m-%d" -j "2012-02-29" >/dev/null 2>&1
The -fprovides the input format, and the -jtells dateto simply output the date, not attempt to set the system clock.
在-f提供了输入格式，并-j讲述date简单地输出日期，不要试图设置系统时钟。
回答by y?s??la
I came up with this little function:
我想出了这个小功能：
function isDateInvalid()
{
date -d "" "+%m/%d/%Y" > /dev/null 2>&1
res=$?
echo "$res"
}
isDateInvalid "2012-02-219"
1
isDateInvalid "2012-02-29"
0
回答by deadElk
for y in {2013..2014}; do
for m in {01..12}; do
for d in {01..31}; do
[[ ! "`date -jf %Y%m%d $y$m$d +%Y%m%d`" = "$y$m$d" ]] && continue
echo $y.$m.$d
done
done
done
if strings match, loop will proceed ...
如果字符串匹配，循环将继续......
