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 $2}'
Note:
^* was replaced with ^\* so as to ensure that * is matched as a literal, and (.*) was replaced with .+, because (a) there is nothing to be gained by a capture group here, and (b) it's fair to assume that at least one letter follows the --.
-F '\\* |:' uses either literal *<space> or : as the field separator, which ensures that only the --... token (the second field) is printed.
Related
I have a string "12G 39G 24% /dev" . I have to extract the value '24'. I have used the below regex
grep '[0-9][0-9]%' -o
But I am getting output as 24%. I want only 24 as output and don't want '%' character. How to modify the regex script to extract only 24 as value?
One option would be to just grep again for the digits:
grep -o '[0-9][0-9]%' | grep -o '[0-9][0-9]'
However, if you want to accomplish this with a single regex, you can use the following:
grep -Po '[0-9]{2}(?=%)'
Note the -P option in this case; vanilla grep doesn't seem to support the (?=%) "look-around" part.
The most common way not to capture something is using look-around assertions:
Use it like this
grep -oP '[0-9][0-9](?=%)'
It's worth noting that GNU grep support the -P option to enable Perl compatible regex syntax, however it is not included with OS X. On Linux, it will be available by default. A workaround would be to use ack instead.
But I'd still recommend to use GNU grep on OS X by default. It can be installed on OSX using Homebrew with the command brew grep install
Also, see How to match, but not capture, part of a regex?
You can use sed as an alternative:
sed -rn 's/(^.*)([[:digit:]]{2})(%.*$)/\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.
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.)
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 am trying to use sed to get some info that is encoded within the path of a file which is passed as a parameter to my script (Bourne sh, if it matters).
From this example path, I'd like the result to be 8
PATH=/foo/bar/baz/1-1.8/sing/song
I first got the regex close by using sed as grep:
echo $PATH | sed -n -e "/^.*\/1-1\.\([0-9][0-9]*\).*/p"
This properly recognized the string, so I edited it to make a substitution out of it:
echo $PATH | sed -n -e "s/^.*\/1-1\.\([0-9][0-9]*\).*/\1/"
But this doesn't produce any output. I know I'm just not seeing something simple, but would really appreciate any ideas about what I'm doing wrong or about other ways to debug sed regular expressions.
(edit)
In the example path the components other than the numerical one can contain numbers similar to the numeric path component that I listed, but not quite the same. I'm trying to exactly match the component that that is 1-1. and see what some-number is.
It is also possible to have an input string that the regular expression should not match and should product no output.
The -n option to sed supresses normal output, and since your second line doesn't have a p command, nothing is output. Get rid of the -n or stick a p back on the end
It looks like you're trying to get the 8 from the 1-1.8 (where 8 is any sequence of numerics), yes? If so, I would just use:
echo /foo/bar/baz/1-1.8/sing/song | sed -e "s/.*\/1-1\.//" -e "s/[^0-9].*//"
No doubt you could get it working with one sed "instruction" (-e) but sometimes it's easier just to break it down.
The first strips out everything from the start up to and including 1-1., the second strips from the first non-numeric after that to the end.
$ echo /foo/bar/baz/1-1.8/sing/song | sed -e "s/.*\/1-1\.//" -e "s/[^0-9].*//"
8
$ echo /foo/bar/baz/1-1.752/sing/song | sed -e "s/.*\/1-1\.//" -e "s/[^0-9].*//"
752
And, as an aside, this is actually how I debug sed regular expressions. I put simple ones in independent instructions (or independent part of a pipeline for other filtering commands) so I can see what each does.
Following your edit, this also works:
$ echo /foo/bar/baz/1-1.962/sing/song | sed -e "s/.*\/1-1\.\([0-9][0-9]*\).*/\1/"
962
As to your comment:
In the example path the components other than the numerical one can contain numbers similar to the numeric path component that I listed, but not quite the same. I'm trying to exactly match the component that that is 1-1. and see what some-number is.
The two-part sed command I gave you should work with numerics anywhere in the string (as long as there's no 1-1. after the one you're interested in). That's because it actually deletes up to the specific 1-1. string and thereafter from the first non-numeric). If you have some examples that don't work as expected, toss them into the question as an update and I'll adjust the answer.
You can shorten you command by using + (one or more) instead of * (zero or more):
sed -n -e "s/^.*\/1-1\.\([0-9]\+\).*/\1/"
don't use PATH as your variable. It clashes with PATH environment variable
echo $path|sed -e's/.*1-1\.//;s/\/.*//'
You needn't divide your patterns with / (s/a/b/g), but may choose every character, so if you're dealing with paths, # is more useful than /:
echo /foo/1-1.962/sing | sed -e "s#.*/1-1\.\([0-9]\+\).*#\1#"
My GPS logger occassionally leaves "unfinished" lines at the end of the log files. I think they're only at the end, but I want to check all lines just in case.
A sample complete sentence looks like:
$GPRMC,005727.000,A,3751.9418,S,14502.2569,E,0.00,339.17,210808,,,A*76
The line should start with a $ sign, and end with an * and a two character hex checksum. I don't care if the checksum is correct, just that it's present. It also needs to ignore "ADVER" sentences which don't have the checksum and are at the start of every file.
The following Python code might work:
import re
from path import path
nmea = re.compile("^\$.+\*[0-9A-F]{2}$")
for log in path("gpslogs").files("*.log"):
for line in log.lines():
if not nmea.match(line) and not "ADVER" in line:
print "%s\n\t%s\n" % (log, line)
Is there a way to do that with grep or awk or something simple? I haven't really figured out how to get grep to do what I want.
Update: Thanks #Motti and #Paul, I was able to get the following to do almost what I wanted, but had to use single quotes and remove the trailing $ before it would work:
grep -nvE '^\$.*\*[0-9A-F]{2}' *.log | grep -v ADVER | grep -v ADPMB
Two further questions arise, how can I make it ignore blank lines? And can I combine the last two greps?
The minimum of testing shows that this should do it:
grep -Ev "^\$.*\*[0-9A-Fa-f]{2}$" a.txt | grep -v ADVER
-E use extended regexp
-v Show lines that do not match
^ starts with
.* anything
\* an asterisk
[0-9A-Fa-f] hexadecimal digit
{2} exactly two of the previous
$ end of line
| grep -v ADVER weed out the ADVER lines
HTH, Motti.
#Motti's answer doesn't ignore ADVER lines, but you easily pipe the results of that grep to another:
grep -Ev "^\$.*\*[0-9A-Fa-f]{2}$" a.txt |grep -v ADVER
#Tom (rephrased) I had to remove the trailing $ for it to work
Removing the $ means that the line may end with something else (e.g. the following will be accepted)
$GPRMC,005727.000,A,3751.9418,S,14502.2569,E,0.00,339.17,210808,,,A*76xxx
#Tom And can I combine the last two greps?
grep -Ev "ADVER|ADPMB"
#Motti: Combining the greps isn't working, it's having no effect.
I understand that without the trailing $ something else may folow the checksum & still match, but it didn't work at all with it so I had no choice...
GNU grep 2.5.3 and GNU bash 3.2.39(1) if that makes any difference.
And it looks like the log files are using DOS line-breaks (CR+LF). Does grep need a switch to handle that properly?
#Tom
GNU grep 2.5.3 and GNU bash 3.2.39(1) if that makes any difference.
And it looks like the log files are using DOS line-breaks (CR+LF). Does grep need a switch to handle that properly?
I'm using grep (GNU grep) 2.4.2 on Windows (for shame!) and it works for me (and DOS line-breaks are naturally accepted) , I don't really have access to other OSs at the moment so I'm sorry but I won't be able to help you any further :o(