regex - bash, extract string before a colon

Question

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.

Answer 1

cut -d: -f1

or

awk -F: '{print $1}'

or

sed 's/:.*//'