Reputation: 37
In Linux (Raspbian distro) I'm trying to extract the date part of a file name which works when I type it in directly into the terminal (see below).
$ file1="access_point20140821.csv"
$ echo $file1 | cut -c13-20
$ 20140821
However, when I put this into a shell script I can't seem to extract the date part of the file name. The echo line just returns "Date Part" with nothing following it. I suspect this is something to do with how I am assigning the variable DATE_PART. Can anyone help?
FILENAME="access_point20140821.csv"
DATE_PART=$FILENAME | cut -c13-20
echo "Date Part $DATE_PART"
Upvotes: 0
Views: 2303
Reputation: 31494
You are not echoing the file, you are trying to execute it, you should do:
FILENAME="access_point20140821.csv"
DATE_PART=$(echo $FILENAME | cut -c13-20)
echo "Date Part $DATE_PART"
Upvotes: 1