How to use sed on Linux to get values from file name?

Question

how to retrive every portion separately from following file name? DSA4020_frontcover_20346501_2011-05.doc

I want to retrieve informations as below; name = DSA4020

type = frontcover

id = 20346501

date = 2011-05

is it possible to do with sed??

Answer 1

How about this 'sed' recipe.

echo "DSA4020_frontcover_20346501_2011-05.doc"  \
 | sed 's|\([^_]*\)_\([^_]*\)_\([^_]*\)_\([^\.]*\).*|name=\1,type=\2,id=\3,date=\4|'

gives this nice CSV form,

name=DSA4020,type=frontcover,id=20346501,date=2011-05

Answer 2

You can use the following to split at every _ after removing the extension:

$ echo "DSA4020_frontcover_20346501_2011-05.doc" | cut -d. -f-1 | cut -d_ -f1
DSA4020

Replace the very last digit by 2, 3, 4 to get the individual value each time.

Answer 3

If the number of fields is constant :

for i in 1 2 3 4 5; do  
    VARS[${i}]=$(echo ${yourfilename} | cut -d _ -f${i})
done

Then access ${VARS[i]}...

Alternative :

VARS[${i}]=$(echo ${yourfilename} | awk -v i=${i} -F_ '{print $i}'

And if supported (bash 3+)

for i in {1..5}; do (...)

---

If you have multiple filenames, just add a for loop to skim through all of them.

Answer 4

Split the filename and store in positional parameters:

set -- $(sed 's/_/ /g' <<< ${filename%.doc})
name=$1
type=$2
id=$3
date=$4