我有一个这种格式的日期:“27 JUN 2011”，我想将它转换为 20110627
可以用 bash 做吗？
最佳答案
#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
More on Dates
请注意，这仅适用于 GNU 日期
我读过:
关于linux - 在 bash 中转换日期格式，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/6508819/