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I have a text file on Linux where the contents are like below:
help.helloworld.com:latest.world.com
dev.helloworld.com:latest.world.com
I want to get the contents before the colon like below:
help.helloworld.com
dev.helloworld.com
How can I do that within the terminal?
This is what cut is for:
$ cat file
help.helloworld.com:latest.world.com
dev.helloworld.com:latest.world.com
foo:baz:bar
foo
$ cut -d: -f1 file
help.helloworld.com
dev.helloworld.com
foo
foo
You just set the delimiter to : with -d: and tell it to only print the 1st field (-f1).
Or an alternative:
$ grep -o '^[^:]*' file
help.helloworld.com
dev.helloworld.com
This returns any characters beginning at the start of each line (^) which are no colons ([^:]*).
Would definitely recommend awk:
awk -F ':' '{print $1}' file
Uses : as a field separator and prints the first field.
Considering the following file file.txt:
help.helloworld.com:latest.world.com
dev.helloworld.com:latest.world.com
no.colon.com
colon.at.the.end.com:
You can use sed to remove everything after the colon:
sed -e 's/:.*//' file.txt
This works for all the corner cases pointed out in the comments—if it ends in a colon, or if there is no colon, although these weren't mentioned in the question itself. Thanks to @Rakesh Sharma, @mirabilos, and @Freddy for their comments. Answering questions is a great way to learn.
if in the s///p syntax, you need to modify your regex to take care of lines with no colons, something like, sed -nEe 's/([^:]*)(:.*|)/\1/p'. Note this requires GNU sed but since anyway you are on GNU sed so this shouldn't matter.
Aug 28, 2019 at 5:05
sed -n '/:/s/^\([^:]*\):.*$/\1/p (add --posix if you use GNU sed, just to spite the extensionism of theirs)
Aug 28, 2019 at 18:09
Requires GNU grep. It would not work with the default grep on e.g. macOS or any of the other BSDs.
Do you mean like this:
grep -oP '.*(?=:)' file
Output:
help.helloworld.com
dev.helloworld.com
echo foo:bar:baz | grep -oP '.*(?=:)'. This will work for the OP's example, but not for the general case as described in the question.
You could achieve this with bash string handling, by removing the longest match from the string directly for each line read like so:
for line in $(cat inputfile); do echo "${line%%:*}"; done
This might be a useful alternative if you are parsing the file in a shell script (though I suspect using cut might be more efficient).
In pure POSIX shell without using external commands, I'd do:
#/bin/sh
IFS=:
while read -r a _; do
echo "$a"
done < file.txt
unset IFS
greputility is used for looking for lines matching regular expressions. You could possibly use it here, but it would be more appropriate to use a tool that extracts data from fields given some delimiter, such as thecututility.grepis the right tool to solve the actual problem.