regexstringbashsedsubstring

bash, extract string before a colon

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.

Solution 
  • cut -d: -f1

    or 

    awk -F: '{print $1}'

    or 

    sed 's/:.*//'