Extracting a date from a filename and converting it in Bash-CodePudding

I'm trying to extract the date from a filename and assign the converted date to a variable.

E.g.

Filename="SomeFilename 05 May 2022.zip"

And I want to extract 05052022 and assign that to a variable.

CodePudding user response：

The following method is a little lengthy but you can use a bash regex for extracting the info and a case switch for converting the month:

#!/bin/bash

filename='SomeFilename 05 May 2022.zip'

if [[ $filename =~ ([0-9]{2})\ ([A-Z][a-z]{2})\ ([0-9]{4})\.[^.] $ ]]
then
    d=${BASH_REMATCH[1]}

    case ${BASH_REMATCH[2]} in
    Jan) m=01;; Feb) m=02;; Mar) m=03;;
    Apr) m=04;; May) m=05;; Jun) m=06;;
    Jul) m=07;; Aug) m=08;; Sep) m=09;;
    Oct) m=10;; Nov) m=11;; Dec) m=12;;
    esac

    y=${BASH_REMATCH[3]}

    date=$d$m$y
fi

CodePudding user response：

Using regex and grep (PCRE mode) the date command:

Filename="SomeFilename 05 May 2022.zip"
date=$(grep -oP '\d{2}\s \w \s \d{4}(?=\.zip)' <<< "$Filename")
res=$(date -d "$date" ' %d%m%Y')

CodePudding user response：

Using sed

$ Filename="SomeFilename 05 May 2022.zip"
$ date=$(sed s"/[^ ]* \([^.]*\).*/date -d '\1'  '%d%m%Y'/e" <<< $Filename)
$ echo "$date"
05052022