I am writing a bash script to extract a domain out of string of the form:
....domain name example.com...
and I want to output example.com.
How can I use matching to output this domain name?
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3The question is too loosely defined to be answered properly. Try to think of some corner cases and describe how they should be handled.tripleee– tripleee2012-11-05 21:08:45 +00:00Commented Nov 5, 2012 at 21:08
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4 Answers
You should better define the pattern of your domain in order to better match it in the string. After that you can use grep to match a regex describing your domain pattern:
domain=`echo "....domain name xxx.com..." | grep -om 1 -G "[^ ]*\.com"`
Comments
Without assuming the domain name ends in ".com"
grep -oP '(?<=domain name )\S+'
meaning: look for a sequence of non-whitespace characters following the string "domain name "
Comments
with GNU grep
echo "random words domain name example.com random words" | grep -oP "domain name \K[^ ]+\.com"
Comments
If the the format is exactly as you say, then this will suffice:
awk '/domain name/{print $3}'
If the string is stored in a variable, you can use it as follows:
awk '/domain name/{print $3}' <<< $string_name
If the string is stored in a file with other strings, one per line:
awk '/domain name/{print $3}' < input_file > output_file
4 Comments
user1190650
Thanks so much. I'm just wondering- in your second example, how does $3 tell awk to output the word that is directly following "domain name"?
tripleee
It assumes they are always the first two words.
user1190650
ok, is there a way to get the word directly after "domain name" if there are words prior (i.e. "... ... abc def domain name xxx.com....")?
sampson-chen
@user1190650 there is, but it would be easier for everyone if you more explicitly defined the requirements of the problem in your original question.