Read filename as a variable

kicap · 08-11-2009, 11:07 PM

I am really2 appreciate if someone could gave me some idea. I have a file name '2009-08-06_00-00-00.csv' and how could i produce an output as '2009-08-06' that yield from the given filename by using shell script?.

Thanks

tuxdev · 08-12-2009, 12:10 AM

If you care about the first 10 characters:

Code:

echo $FILENAME | cut -b 1-10

If you care about everything before the _

Code:

echo $FILENAME | sed s/_.*//

ArfaSmif · 08-12-2009, 02:10 AM

basename file.csv .csv

will give you the correct result (man basename)
oops didn't read the question properly did I?

David the H. · 08-12-2009, 04:27 PM

Using no eternal tools, only bash built-ins:

Code:

$ filename="2009-08-06_00-00-00.csv"

$ echo ${filename%_*}  #remove everything from the last _ to the end
2009-08-06

$ echo ${filename#*_}  #remove everything from the beginning to the first _
00-00-00.csv

$ echo ${filename%.*}  #remove everything from the last . to the end
2009-08-06_00-00-00

$ echo ${filename#*.}  #remove everything from the beginning to the first .
csv

$ echo ${filename%%-*} #remove everything from the first - to the end
2009

$ echo ${filename##*-} #remove everything from the beginning to the last -
00.csv

$ echo ${filename:5:2}  #from position 5, print 2 characters
08

$ echo ${filename/00/11}  #replace the first 00 with 11
2119-08-06_00-00-00.csv

$ echo ${filename/_00/_11}  #replace the first _00 with _11
2009-08-06_11-00-00.csv

$ echo ${filename//-/_}  #change all - to _
2009_08_06_00_00_00.csv

Parameter substitution is a wonderful thing.