If I have a file with rows like this
/some/random/file.csv:some string
/some/random/file2.csv:some string2
Is there some way to get a file that only has the first part before the colon, e.g.
/some/random/file.csv
/some/random/file2.csv
I would prefer to just use a bash one liner, but perl or python is also ok.
cut -d: -f1
or
awk -F: '{print $1}'
or
sed 's/:.*//'