我有这样的日期格式:“ 2011年6月27日”,我想将其转换为20110627
可以用bash做吗?
#since this was yesterday date -dyesterday +%Y%m%d #more precise, and more recommended date -d'27 JUN 2011' +%Y%m%d #assuming this is similar to yesterdays `date` question from you #http://stackoverflow.com/q/6497525/638649 date -d'last-monday' +%Y%m%d #going on @seth's comment you could do this DATE="27 jun 2011"; date -d"$DATE" +%Y%m%d #or a method to read it from stdin read -p " Get date >> " DATE; printf " AS YYYYMMDD format >> %s" `date -d"$DATE" +%Y%m%d` #which then outputs the following: #Get date >> 27 june 2011 #AS YYYYMMDD format >> 20110627 #if you really want to use awk echo "27 june 2011" | awk '{print "date -d\""$1FS$2FS$3"\" +%Y%m%d"}' | bash #note | bash just redirects awk's output to the shell to be executed #FS is field separator, in this case you can use $0 to print the line #But this is useful if you have more than one date on a line
有关日期的更多信息
请注意,这仅适用于GNU日期
我读过:
-d可以通过替换sunfreeware.com版本的日期来解决无法支持的Solaris版本的日期
-d