bash, extract string before a colon

bash, regex, sed, string, substring

Solution

cut -d: -f1

or

awk -F: '{print $1}'

or

sed 's/:.*//'

Problem

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.

Original source