How to convert a bash string to date format?-CodePudding

I have a bash string 20220416124334 (the string is not epoch) and I want to convert it to date format so it should look like this: 2022/04/16 12:43:34

I've tried:

[manjaro@manjaro ~]$ date -d '20220416124334' " %Y/%m/%d %H:%M:%S"
date: invalid date ‘20220416124334’

How should I do it correctly ?

CodePudding user response：

I would expect to use the date command, as you employ it, with a number that represents the number of seconds since the epoch. It seems your number is a formatted date yyyymmddhhmiss.

If you don't need to validate the string that represents a formatted date, then you can use sed to insert extra formatting characters:

echo '20220416124334' | sed -E 's/(....)(..)(..)(..)(..)(..)/\1\/\2\/\3 \4:\5:\6/'

If you end up taking as input the number of seconds since the epoch, then do it this way:

date -r 1650074146 " %Y/%m/%d %H:%M:%S"