Extracting date pattern in shell script

Question

I am trying to extract a pattern in shell script.I have the below variable with date format(yyyyMMdd) has value.

field=20200825

Here i want to extract the date,month,year. In case of months if 0 is prefixed, for e.g if 08 is there i need only 8 to be extracted from it.

Can anybody help me on this?

So https://stackoverflow.com/questions/428109/extract-substring-in-bash ? — KamilCuk, Aug 25 '20 at 08:12

score 0 · Answer 1 · answered Aug 26 '20 at 12:55

An example script, extracts y/m/d info from string based on offset + length, and trim a possible leading '0' from month

#!/bin/bash

field=20200825
year=${field:0:4}
month=${field:4:2}
month=${month#0}
day=${field:6:2}

echo year $year
echo month $month
echo day $day

score 0 · Answer 2 · answered Aug 26 '20 at 13:09

As @Milag answered, you can use Substring Expansion to cut your needed variables. This is true because you said in a comment that you are using bash.

If you need a more portable way to do it, you can (among other solutions) use printf formatting:

#!/bin/sh
field=20200826
year="$(printf "%.4s" "${field}")"
field="${field#????}"
month="$(printf "%.2s" "${field}")"
month="${month#0}"
day="${field#??}"
printf 'year: %s\n' "${year}"
printf 'month: %s\n' "${month}"
printf 'day: %s\n' "${day}"

Extracting date pattern in shell script

2 Answers2