Shell (Bash) parameter expansion expressions

Question

I have some code that converts a date in a file name into a path.

#!/bin/bash
aFile="bla_2016-11-24_17-24.zip" # Many files, each with same format: "bla_{year}-{month}-{day}_{hour}-{minute}.zip"
aDate=${aFile#*_} # remove "bla_"
aDate=${aDate%.*} # remove ".zip"
aDate=${aDate//_/ } # replace "_" with " "
aDate=(${aDate//-/ }) # replace "-" with " " and make array
aPath="${aDate[0]}/${aDate[1]}/${aDate[2]}/${aDate[3]}/${aDate[4]}"
mkdir -p $aPath

I've seen both of these (1, 2), which suggest that nested parameter expansion expressions in bash scripting are not possible.

I was wondering if there might be a better way to implement my code above, more specifically reduce the length of the code. It's just five lines, so not a huge irritation if it's not possible, but it seems like I'm doing it in a dumb way.

Python solution:

aFile = "bla_2016-11-24_17-24.zip"

import re
aPath = "/".join(re.split("[_-]", aFile[4:-4])) # 2016/11/24/17/24

With GNU sed: `echo "bla_2016-11-24_17-24.zip" | sed -E 's/^[^_]+_(.*)\.zip$/\1/;s|[_-]|/|g'` Output: `2016/11/24/17/24` or with awk: `echo "bla_2016-11-24_17-24.zip" | awk -F '[_.-]' '{OFS="/"; print $2,$3,$4,$5,$6}'` — Cyrus, May 07 '17 at 19:21

score 2 · Accepted Answer · answered May 07 '17 at 21:13

2

With bash:

aFile="bla_2016-11-24_17-24.zip"
[[ ${aFile//[_.-]//} =~ ^[^/]*/(.*)/[^/]*?$ ]] && echo "${BASH_REMATCH[1]}"

Output:

2016/11/24/17/24

answered May 07 '17 at 21:13

Cyrus

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I like this more than @RicardoBranco solution because it doesn't reference the array 5 times – dǝɥɔS ʇoıןןƎ May 09 '17 at 12:29
I like it too. Didn't know about this BASH_REMATCH feature. Upvote. – Ricardo Branco May 09 '17 at 17:28

Ricardo Branco · Answer 2 · 2017-05-09T17:53:53.967

1

Here's a quicker Bash solution:

aFile="bla_2016-11-24_17-24.zip"
arr=($(echo ${aFile//[_-\.]/ }))
aPath=${arr[1]}/${arr[2]}/${arr[3]}/${arr[4]}/${arr[5]}

EDIT: This solution only works with recent versions of Bash such as the one shipped with Ubuntu 16.04 (4.3.46). It fails with 4.3.30 (shipped by Debian 8).

edited May 09 '17 at 17:53

answered May 07 '17 at 21:11

Ricardo Branco

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The problem is with the version of Bash. With version 4.1.5 (from the Docker image of Debian 6) it fails. My version is 4.3.46 (from Ubuntu 16.04). I fear we must debug our bashisms on older versions... :( Which version is yours? – Ricardo Branco May 09 '17 at 17:45
I was thinking the same. On MacOS though so my version was `3.2.57` or something similar. I updated, now I have `4.4.12`. I guess the latest version is too high? :o – dǝɥɔS ʇoıןןƎ May 09 '17 at 17:58
I think it's better than having that old version. Ubuntu 17.04 uses 4.4.5 and the upcoming Debian 9 4.4.11. – Ricardo Branco May 09 '17 at 18:03

score 0 · Answer 3 · answered May 07 '17 at 20:41

Actually, I solved this. I tried a solution with sed then realised that's dumb; I could just use my Python solution. So for anyone wondering, here's mine:

aFile="bla_2016-11-24_17-24.zip"
aPath="$(python -c 'import sys, re; print "/".join(re.split("[_-]", sys.argv[1][4:-4]))' "$aFile")" # 2016/11/24/17/24

Shell (Bash) parameter expansion expressions

3 Answers3