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

User · Answer

This has been asked so many times so that a user with over 1000 points ask for this is some strange
But just to show just another way to do it:

echo "/some/random/file.csv:some string" | awk '{sub(/:.*/,x)}1'
/some/random/file.csv

User · Answer

Try this in pure bash   FRED   some random file csv some string  a   FRED     echo  a   Here is some documentation that helps

User · Answer

Another pure Bash solution   while IFS     read a b   do   echo   a  done  lt    infile   gt    outfile

User · Answer

This works for me you guys can try it out INPUT  ubuntu x 1000 1000 Ubuntu  home ubuntu  bin bash  SUBSTRING   echo  INPUT  cut -d  -f1  echo  SUBSTRING

User · Answer

cut -d  -f1   or   awk -F    print  1     or   sed  s

User · Answer

Another pure BASH way    gt  s   some random file csv some string   gt  echo    s        some random file csv

[regex] bash, extract string before a colon

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