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Get Substring from file name.

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# 8  
Old 03-17-2013
If the files contain 1 extra digit (for instance a version number somewhere) then most of these methods will not render the correct results. You could select on the basis of how many digits need to be minimally present, for example 4:
Code:
IFS=_.
for file in *_*[0-9][0-9][0-9][0-9]*_*.*
do
  set -- $file
  for i do
    case $i in (*[0-9][0-9][0-9][0-9]*)
      printf "%d\n" "$i"
    esac
  done
done
IFS=$oldIFS

or
Code:
IFS=_.
for file in *_*[0-9][0-9][0-9][0-9]*_*.*
do
  for i in $file
  do
    case $i in (*[0-9][0-9][0-9][0-9]*)
      printf "%d\n" "$i"
    esac
  done
done
IFS=$oldIFS

Or:
Code:
ls | sed -n 's/.*_\([0-9]\{4,\}\)_.*\..*/\1/p'

Last edited by Scrutinizer; 03-17-2013 at 06:50 AM..
This User Gave Thanks to Scrutinizer For This Post:
pinnacle
# 9  
Old 03-17-2013
Quote:
Originally Posted by hanson44
Here is one way to extract the number from those files:
Code:
 for i in *; do
  j=`echo $i | sed 's/.*_\([0-9]\+\).*/\1/'`
  echo $j
done

Code:
 for i in *; do
  j=`echo $i | sed -r 's/.*_([0-9]+).*/\1/'`
  echo $j
done

Because * is 0 or more, _[0-9]* was picking up the first '_' instead of the one before '2'. And you have to use \1 to do the replacement.
The above ones does not seem to be working. Please see below:

Code:
echo $i | sed 's/.*_\([0-9]\+\).*/\1/'
adio_idfl_201302_df.txt


Code:
echo $i | sed -r 's/.*_([0-9]+).*/\1/'
sed: illegal option -- r
Usage:  sed [-n] [-u] Script [File ...]
        sed [-n] [-u] [-e Script] ... [-f Script_file] ... [File ...]

Thanks for the help though.

---------- Post updated at 11:36 AM ---------- Previous update was at 11:33 AM ----------

Quote:
Originally Posted by elixir_sinari
You don't need sed to do such a simple task (provided your shell is capable of doing the following):
Code:
i='adio_idfl_201302_df.txt'
 
echo "${i//[!0-9]/}"
201302


Looks like my shell is not capable of doing this.

Code:
"${i//[!0-9]/}": bad substitution

---------- Post updated at 11:38 AM ---------- Previous update was at 11:36 AM ----------

Quote:
Originally Posted by anbu23
Code:
$ ls
adio_idfl_201302_df.txt  dalkd_20130301.csv  dfa_201301_dll.ctl
$ ls | tr -dc [:digit:]
20130220130301201301
$ ls | tr -dc '[\n[:digit:]]'
201302
20130301
201301


Thanks this is working.

---------- Post updated at 11:38 AM ---------- Previous update was at 11:38 AM ----------

Code:
ls | sed -n 's/.*_\([0-9]\{4,\}\)_.*\..*/\1/p'

Thanks I will be using the above code.
# 10  
Old 03-17-2013
Quote:
Originally Posted by pinnacle
The above ones does not seem to be working. Please see below:

Code:
echo $i | sed 's/.*_\([0-9]\+\).*/\1/'
adio_idfl_201302_df.txt

The reason for the failure is that the highlighted portion of the regular expression is a GNU extension.

I see no reason to limit one's options by ever using \+. It may be a bit longer, but \{1,\} works everywhere.

Regards,
Alister
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