How to get Substring from Filename in Unix shellscripting

Question

I want a shell script to get MMDDYYYY from the file with a name as mentioned below file

linuxbox.23566.MMDDYYYYHHMMSS.zip

jaypal singh · Accepted Answer · 2014-03-14 12:18:24Z

4

Using bash string functions:

for file in *.zip; do 
    file="${file%.*}"
    file="${file##*.}"
    echo "${file:0:8}"
done

file="${file%.*}": Gets rid of the extension and stores the new name in file variable
file="${file##*.}": Gets rid of the longest match from beginning and stores the name in file variable
echo "${file:0:8}": echoes the first 8 characters of whats left.

$ ls
linuxbox.23566.MMDDYYYYHHMMSS.zip
$ for file in *; do file="${file%.*}"; file="${file##*.}"; echo "${file:0:8}"; done
MMDDYYYY

answered Mar 14, 2014 at 12:18

jaypal singh

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Comments

fedorqui · Accepted Answer · 2014-03-14 11:49:34Z

1

With cut:

$ cut -d. -f3 <<< "linuxbox.23566.MMDDYYYYHHMMSS.zip" | cut -c-8
MMDDYYYY

Because the first part is returning:

$ cut -d. -f3 <<< "linuxbox.23566.MMDDYYYYHHMMSS.zip"
MMDDYYYYHHMMSS

And then it gets the first 8 chars.

answered Mar 14, 2014 at 11:49

fedorqui

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the above is working in terminal. If i'm adding this in shell scripting this is not working. Can you provide shell script for the above command

What did you exactly try? Please update your question showing what you did.