CODEWITHSUNDEEP
bashparsingtokenize
pufos
pufos

Reputation: 2930

bash parse filename

Is there any way in bash to parse this filename :

$file = dos1-20120514104538.csv.3310686

into variables like $date = 2012-05-14 10:45:38 and $id = 3310686 ?

Thank you

Upvotes: 6

Views: 14140

Answers (4)

user unknown
user unknown

Reputation: 36250

Extract id:

f='dos1-20120514104538.csv.3310686'
echo ${f/*./}
# 3310686
id=${f/*./}

Remove prefix, and extract core date numbers:

noprefix=${f/*-/}
echo ${noprefix/.csv*/}
# 20120514104538
ds=${noprefix/.csv*/}

format the date like this (only partially done:)

echo $ds | sed -r 's/(.{4})(.{2})(.{2})/\1.\2.\3/'

You can alternatively split the initial variable into an array, 

echo $f
# dos1-20120514104538.csv.3310686

after exchanging - and . like this:

echo ${f//[-.]/ }
# dos1 20120514104538 csv 3310686

ar=(${f//[-.]/ })
echo ${ar[1]}
# 20120514104538

echo ${ar[3]}
# 3310686

The date transformation can be done via an array similarly:

dp=($(echo 20120514104538  | sed -r 's/(.{2})/ \1/g'))
echo ${dp[0]}${dp[1]}-${dp[2]}-${dp[3]} ${dp[4]}:${dp[5]}:${dp[6]}

It splits everything into groups of 2 characters:

echo ${dp[@]}
# 20 12 05 14 10 45 38

and merges 2012 together in the output.

Upvotes: 3

kojiro
kojiro

Reputation: 77137

All of this can be done with Parameter Expansion. Please read about it in the bash manpage.

$ file='dos1-20120514104538.csv.3310686'
$ date="${file#*-}" # Use Parameter Expansion to strip off the part before '-'
$ date="${date%%.*}" # Use PE again to strip after the first '.'
$ id="${file##*.}" # Use PE to get the id as the part after the last '.'
$ echo "$date"
20120514104538
$ echo "$id"
3310686

Combine PEs to put date back together in a new format. You could also parse the date with GNU date, but that would still require rearranging the date so it can be parsed. In its current format, this is how I would approach it:

$ date="${date:0:4}-${date:4:2}-${date:6:2} ${date:8:2}:${date:10:2}:${date:12:2}"
$ echo "$date"
2012-05-14 10:45:38

Upvotes: 17

Ozair Kafray
Ozair Kafray

Reputation: 13539

You can tokenize the string first for - and then for .. There are various threads on SO on how to do this:

  1. How do I split a string on a delimiter in Bash?
  2. Bash: How to tokenize a string variable?

To transform 20120514104538 into 2012-05-14 10:45:38 :

Since we know that first 4 characters is year, next 2 is months and so on, you will first need to break this token into sub-strings and then recombine into a single string. You can start with the following answer:

  1. https://stackoverflow.com/a/428580/365188

Upvotes: 1

Dennis Williamson
Dennis Williamson

Reputation: 360315

Using Bash's regular expression feature:

file='dos1-20120514104538.csv.3310686'
pattern='^[^-]+-([[:digit:]]{4})'
for i in {1..5}
do
    pattern+='([[:digit:]]{2})'
done
pattern+='\.[^.]+\.([[:digit:]]+)$'
[[ $file =~ $pattern ]]
read -r _ Y m d H M S id <<< "${BASH_REMATCH[@]}"
date="$Y-$m-$d $H:$M:$S"
echo "$date"
echo "$id"

Upvotes: 3

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