I have a string "12G 39G 24% /dev" . I have to extract the value '24'. I have used the below regex
grep '[0-9][0-9]%' -o
But I am getting output as 24%. I want only 24 as output and don't want '%' character. How to modify the regex script to extract only 24 as value?
One option would be to just grep again for the digits:
grep -o '[0-9][0-9]%' | grep -o '[0-9][0-9]'
However, if you want to accomplish this with a single regex, you can use the following:
grep -Po '[0-9]{2}(?=%)'
Note the -P option in this case; vanilla grep doesn't seem to support the (?=%) "look-around" part.
The most common way not to capture something is using look-around assertions:
Use it like this
grep -oP '[0-9][0-9](?=%)'
It's worth noting that GNU grep support the -P option to enable Perl compatible regex syntax, however it is not included with OS X. On Linux, it will be available by default. A workaround would be to use ack instead.
But I'd still recommend to use GNU grep on OS X by default. It can be installed on OSX using Homebrew with the command brew grep install
Also, see How to match, but not capture, part of a regex?
You can use sed as an alternative:
sed -rn 's/(^.*)([[:digit:]]{2})(%.*$)/\2/p' <<< "12G 39G 24% /dev"
Enable regular expressions with -r or -E and then split the line into 3 sections represented through parenthesis. Substitute the line for the second section only and print.
Use awk:
awk '{print $3+0}'
The value you seek is in the third field, and adding a zero coerces the string to a number, so % is removed.
Related
I understand that what I'm asking can be accomplished using awk or sed, I'm asking here how to do this using GREP.
Given the following input:
.bash_profile
.config/ranger/bookmarks
.oh-my-zsh/README.md
I want to use GREP to get:
.bash_profile
.config/
.oh-my-zsh/
Currently I'm trying
grep -Po '([^/]*[/]?){1}'
Which results in output:
.bash_profile
.config/
ranger/
bookmarks
.oh-my-zsh/
README.md
Is there some simple way to use GREP to only get the first matched string on each line?
I think you can grep non / letters like:
grep -Eo '^[^/]+'
On another SO site there is another similar question with solution.
You don't need grep for this at all.
cut -d / -f 1
The -o option says to print every substring which matches your pattern, instead of printing each matching line. Your current pattern matches every string which doesn't contain slashes (optionally including a trailing slash); but it's easy to switch to one which only matches this pattern at the beginning of a line.
grep -o '^[^/]*' file
Notice the addition of the ^ beginning of line anchor, and the omission of the -P option (which you were not really using anyway) as well as the silly beginner error {1}.
(I should add that plain grep doesn't support parentheses or repetitions; grep -E would support these constructs just fine, of you could switch to toe POSIX BRE variation which requires a backslash to use round or curly parentheses as metacharacters. You can probably ignore these details and just use grep -E everywhere unless you really need the features of grep -P, though also be aware that -P is not portable.)
I'm trying to parse a command's help file to grab all the arguments the command excepts.
Here is some text from the help file:
* --digest:
Set the digest for fingerprinting (defaults to the digest used when
signing the cert). Valid values depends on your openssl and openssl ruby
extension version.
* --debug:
Enable full debugging.
* --help:
Print this help message
* --verbose:
Enable verbosity.
* --version:
Print the puppet version number
I want to just grab --argument and nothing else.
I almost got it with this command, but its still including the ":" which I want to exclude:
puppet cert --help | egrep '^* --(.*):$' | awk '{print $2}'
--all:
--allow-dns-alt-names:
--digest:
--debug:
--help:
--verbose:
--version:
Why is '^* --(.*):$' including the ":" shouldn't it be matching everything between '^* --' and ':$' ?
shouldn't it be matching everything between ^* -- and :$ ?
Actually, no. You're capturing a group, but it won't print just the group. I suggest using the -P flag to use Perl regex, and look arounds. In your case, this might be enough:
$ cert --help | grep -Po '^\* \K--\w+'
Note that I also used the -o option, to print only the matched content, not the whole line. This eliminates the usage of awk.
A more complete line based on your initial thoughts and more look arounds:
$ cert --help | grep -Po '^\* \K--.*(?=:)'
Edit: as noted in the comments and fine answer by mklement0, this requires GNU grep. You can however do the same with Perl itself, which certainly is probably already installed in your system.
$ cert --help | perl -nle 'print $1 if /^\* (--\w+)/'
This works like a line of code inside a loop. Which is automatically generated by the -nle. -n for the input look, -l for the auto line break, and -e to present the line of code.
The line of Perl code prints the first captured group if the line matches the regex. So it combines ideas from your original solution too.
For a complete POSIX compliant answer, check the answer by mklement0 here in this page.
To provide a POSIX-compliant alternative to sidyll's elegant GNU grep answer (which also explains why the OP's approach didn't work):
Update: Avinash Raj points out in a comment that sed is an option, which indeed allows for a POSIX-compliant single-tool solution: sed allows us to match entire lines of interest and replace them with the contents of a capture group (the part of the line of interest):
puppet cert --help | sed -n 's/^\* \(--.*\):$/\1/p'
Note that since sed is used without the - nonstandard - -r / -E option, a basic regular expression must be used, where ( and ) must be \-escaped to act as capture-group delimiters.
Original answer:
puppet cert --help | egrep '^\* --.+:$' | awk -F '\\* |:' '{print $2}'
Note:
^* was replaced with ^\* so as to ensure that * is matched as a literal, and (.*) was replaced with .+, because (a) there is nothing to be gained by a capture group here, and (b) it's fair to assume that at least one letter follows the --.
-F '\\* |:' uses either literal *<space> or : as the field separator, which ensures that only the --... token (the second field) is printed.
So I run a curl command and grep for a keyword.
Here is the (sanitized) result:
...Dir');">Town / Village</a></th><th>Phone Number</th></tr><tr class="rowodd"><td><a href="javascript:calldialog('ASDF','&Mode=view&helloThereId=42',600,800);"...
I want to get the number 42 - a command line one-liner would be great.
search for the string helloThereId=
extract the number right beside it (42 in the above case)
Does anyone have any tips for this? Maybe some regex for numbers? I'm afraid I don't have enough experience to construct an elegant solution.
You could use grep with -P (Perl-Regexp) parameter enabled.
$ grep -oP 'helloThereId=\K\d+' file
42
$ grep -oP '(?<=helloThereId=)\d+' file
42
\K here actually does the job of positive lookbehind. \K keeps the text matched so far out of the overall regex match.
References:
http://www.regular-expressions.info/keep.html
http://www.regular-expressions.info/lookaround.html
If your grep version supports -P, (as is true for the OP, given that they're on Linux, which comes with GNU grep), Avinash Raj's answer is the way to go.
For the potential benefit of future readers, here are alternatives:
If your grep doesn't support -P, but does support -o, here's a pragmatic solution that simply extracts the number from the overall match in a 2nd step, by splitting the input into fields by =, using cut:
grep -Eo 'helloThereId=[0-9]+' in | cut -d= -f2 file
Finally, if your grep supports neither -P nor -o, use sed:
Here's a POSIX-compliant alternative, using sed with a basic regular expression (hence the need to emulate + with \{1,\} and to escape the parentheses):
sed -n 's/.*helloThereId=\([0-9]\{1,\}\).*/\1/p' file
This will work with any sed on any UNIX OS, even the pre-POSIX default sed on Solaris:
$ sed -n 's/.*helloThereId=\([0-9]*\).*/\1/p' file
42
I have an XML file, the file is made up of one line.
What I am trying to do is extract the "finalNumber" attribute value from the file via Putty. Rather than having to download a copy and search using notepad++.
I've built up a regular expression that I've tested on an On-line Tool, and tried using it within a sed command to duplicate grep functionality. The command runs but doesn't return anything.
RegEx:
(?<=finalNumber=")(.*?)(?=")
sed Command (returns nothing, expected 28, see file extract):
sed -n '/(?<=finalNumber=")(.*?)(?=")/p' file.xml
File Extract:
...argo:finalizedDate="2012-02-09T00:00:00.000Z" argo:finalNumber="28" argo:revenueMonth=""...
I feel like I am close (i could be wrong), am I on the right lines or is there better way to achieve the output?
Nothing wrong with good old grep here.
grep -E -o 'finalNumber="[0-9]+"' file.xml | grep -E -o '[0-9]+'
Use -E for extended regular expressions, and -o to print only the matching part.
Though you already select an answer, here is a way you can do in pure sed:
sed -n 's/^.*finalNumber="\([[:digit:]]\+\)".*$/\1/p' <test
Output:
28
This replaces the entire line by the match number and print (because p will print the entire line so you have to replace the entire line)
This might work for you (GNU sed):
sed -r 's/.*finalNumber="([^"]*)".*/\1/' file
sed does not support look-ahead assertions. Perl does, though:
perl -ne 'print $1 if /(?<=finalNumber=")(.*?)(?=")/'
As I understand, there is no need to use look-aheads here.
Try this one
sed -n '/finalNumber="[[:digit:]]\+"/p'
I would like to get the parameter (without parantheses) of a function call with a regular expression.
I am using egrep in a bash script with cygwin.
This is what I got so far (with parantheses):
$ echo "require(catch.me)" | egrep -o '\((.*?)\)'
(catch.me)
What would be the right regex here?
http://www.greenend.org.uk/rjk/2002/06/regexp.html
What are you looking for - is a lookbehind and lookahead regular expressions.
Egrep cannot do that. grep with perl support can do that.
from man grep:
-P, --perl-regexp
Interpret PATTERN as a Perl regular expression. This is highly experimental and grep -P may warn of unimplemented features.
So
$> echo "require(catch.me)" | grep -o -P '(?<=\().*?(?=\))'
catch.me
If you can use sed then the following would work -
echo "require(catch.me)" | sed 's/.*[^(](\(.*\))/\1/'
You can modify your existing regex to this
echo "require(catch.me)" | egrep -o 'c.*e'
Even though egrep offers this (from the man page)
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
It isn't really the correct utility. SED and AWK are masters at this. You will have much more control using either SED or AWK. :)
From the manual :
grep, egrep, fgrep - print lines matching a pattern
Basically, grep is used to print the complete line, so you won't do anything more.
What you should do is using another tool, maybe perl, for such operations.