[bash] How can I remove all text after a character in bash?

How can I remove all text after a character, in this case a colon (":"), in bash? Can I remove the colon, too? I have no idea how to.

In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed. If you double the character, the longest will be removed.

$ a='hello:world'

$ b=${a%:*}
$ echo "$b"
hello

$ a='hello:world:of:tomorrow'

$ echo "${a%:*}"
hello:world:of

$ echo "${a%%:*}"
hello

$ echo "${a#*:}"
world:of:tomorrow

$ echo "${a##*:}"
tomorrow

Let's say you have a path with a file in this format:

/dirA/dirB/dirC/filename.file

Now you only want the path which includes four "/". Type

$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-4 -d"/"

and your output will be

/dirA/dirB/dirC

The advantage of using cut is that you can also cut out the uppest directory as well as the file (in this example), so if you type

$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-3 -d"/"

your output would be

/dirA/dirB

Though you can do the same from the other side of the string, it would not make that much sense in this case as typing

$ echo "/dirA/dirB/dirC/filename.file" | cut -f2-4 -d"/"

results in

dirA/dirB/dirC

In some other cases the last case might also be helpful. Mind that there is no "/" at the beginning of the last output.

$ echo 'hello:world:again' |sed 's/:.*//'
hello

trim off everything after the last instance of ":"

cat fileListingPathsAndFiles.txt | grep -o '^.*:'

and if you wanted to drop that last ":"

cat file.txt | grep -o '^.*:' | sed 's/:$//'

@kp123: you'd want to replace : with / (where the sed colon should be \/)

egrep -o '^[^:]*:'