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

Answer 1

Try this in pure bash:

FRED="/some/random/file.csv:some string"
a=${FRED%:*}
echo $a

Here is some documentation that helps.

Answer 2

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

Answer 3

Another pure Bash solution:

while IFS=':' read a b ; do
  echo "$a"
done < "$infile" > "$outfile"

Answer 4

Another pure BASH way:

> s='/some/random/file.csv:some string'
> echo "${s%%:*}"
/some/random/file.csv

Answer 5

cut -d: -f1

or

awk -F: '{print $1}'

or

sed 's/:.*//'