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Extracting a portion of the filename

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# 1  
Old 06-18-2013
Extracting a portion of the filename
Hi

I would like to extract the first portion of filename from a list of files.

The filename pattern is of the form 123456789_TEXT_TEXT_TEXT_.csv. I want to extract just the numerical portion of this filename from the list of files and then output this into another text file.

K
# 2  
Old 06-18-2013
Hi,

What have you tried so far ?

Thanks
Pravin
# 3  
Old 06-18-2013
Hi Pravin

I am thinking of using the sed command, but dont know how to go about it.

I have a script

Code:
for i in `ls *TEXTPATTERN*`;do grep String $i | wc -l | grep -v 0 | $i >> matchedfiles.txt;done

on the last $i I would like to just output a portion filenames for matched file.
Last edited by radoulov; 06-18-2013 at 07:53 AM..
# 4  
Old 06-18-2013
Code:
$ a="123456789_TEXT_TEXT_TEXT_.csv"
$ echo ${a%%_*}
123456789
$ echo $a | sed 's/_.*//'
123456789

Code:
for i in *TEXT*
do echo ${i%%_*}
done >matchedfiles.txt

Last edited by ctsgnb; 06-18-2013 at 07:28 AM..
# 5  
Old 06-18-2013
Could this help you ?
This will print the first part of the file (string seperated by "_")

Code:
 
awk -F"_" '{print $1}' ListFiles > newFile.txt

# 6  
Old 06-18-2013
Code:
cut -d"_" -f1 filename

Last edited by radoulov; 06-18-2013 at 05:55 PM..
# 7  
Old 06-18-2013
Linux
This is easieds and simple you can do

Code:
filename="123456789_TEXT_TEXT_TEXT_.csv"
ext="${filename:0:9}"
echo $ext

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