Linux 从日期获取月和日
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Get Month & Day from Date
提问by user1570210
I am trying to get Month and Date from Date in Linux. this is my code
我正在尝试从 Linux 中的日期获取月份和日期。这是我的代码
# Set Date
D="2013/01/17"
# get day
DD=$(D+"%d")
# get day
MM=$(D+"%M")
# Day
echo "Day:"$DD
echo "Month:"$MM
采纳答案by cyfur01
In shor bash:
在sh或bash：
D="2013/01/17"
DAY=$(date -d "$D" '+%d')
MONTH=$(date -d "$D" '+%m')
YEAR=$(date -d "$D" '+%Y')
echo "Day: $DAY"
echo "Month: $MONTH"
echo "Year: $YEAR"
回答by Kent
kent$ D="2013/01/17"
kent$ awk -F/ '{print "year:","Month:","Day:"}'<<<$D
year:2013 Month:01 Day:17
if you want just Month or Day, just leave $2 or $3 there, delete the parts you don't need
如果你只想要月或日，只需在那里留下 2 美元或 3 美元，删除不需要的部分
Edit
编辑
kent$ year=$(awk -F/ '{print }' <<<$D)
kent$ echo $year
2013
回答by Anders Johansson
Or if you want the current date, use date +%Y/%m/%d. If you want them separately you can do something like this:
或者，如果您想要当前日期，请使用date +%Y/%m/%d. 如果你想要它们分开，你可以做这样的事情：
read YYYY MM DD <<<$(date +'%Y %m %d')
echo "Today is Day:$DD Month:$MM"
An easier approach is:
一个更简单的方法是：
DD=$(date +%d)
MM=$(date +%m)
echo "Today is Day:$DD Month:$MM"
However in this case you're executing datetwice, which is inefficient, and if you're really unlucky, the date could change between those two lines ;)
但是，在这种情况下，您执行了date两次，这是低效的，如果您真的不走运，这两行之间的日期可能会改变;)
