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Suppose I have a file with a name ABC_DE_FGHI_10_JK_LMN.csv. I want to extract the ID from the file-name i.e. 10 with the help of ID position and file-name separator. I have following two inputs 

File-name_ID_Position=4; [since 10 is at fourth position in file-name]
File-name_Delimiter="_";

Here ID can be numeric or alpha-numeric. So how extract the 10 from above file with the help of above two inputs. How to achieve this in bash?

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  • Only Bash or is awk usable? Commented Apr 25, 2015 at 18:34
  • It will be OK for me if it returns the desired output and compatible to use in shell script. Commented Apr 25, 2015 at 18:38
  • x="ABC_DE_FGHI_10_JK_LMN.csv"; x=${x#*_*_*_}; echo ${x%_*_*} Commented Apr 25, 2015 at 19:41
  • @Cyrus, PRATIK PATIL changed the file naming convention, and why I deleted my answer, so x="ABC_DE_FGHI_10.csv"; x=${x#*_*_*_}; echo ${x%_*_*} returns 10.csv not just 10 which he wants. However with this marked as a duplicate I guess it doesn't really matter. Commented Apr 25, 2015 at 19:53
  • Using _ or . as separator: x="ABC_DE_FGHI_10.csv"; x=${x#*[_.]*[_.]*[_.]}; echo ${x%[_.]*} Commented Apr 25, 2015 at 22:46

1 Answer 1

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Instead of writing a regex in bash, I would do it with awk:

echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F_ -v pos=4 '{print $pos}'

or if you want the dot to also be a delimiter (requires GNU awk):

echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F'[_.]' -v pos=4 '{print $pos}'
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2 Comments

What if ID comes at last position in file-name? Ex. echo 'ABC_DE_FGHI_10.csv' | awk -F_ -v pos=4 '{print $pos}' then here it returns 10.csv instead of 10
@PRATIKPATIL Then add it to the delimiter (See edit).

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