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 $1 if /(?<=finalNumber=")(.*?)(?=")/'
As I understand, there is no need to use look-aheads here.
Try this one
sed -n '/finalNumber="[[:digit:]]\+"/p'
Related
I have a string "12G 39G 24% /dev" . I have to extract the value '24'. I have used the below regex
grep '[0-9][0-9]%' -o
But I am getting output as 24%. I want only 24 as output and don't want '%' character. How to modify the regex script to extract only 24 as value?
One option would be to just grep again for the digits:
grep -o '[0-9][0-9]%' | grep -o '[0-9][0-9]'
However, if you want to accomplish this with a single regex, you can use the following:
grep -Po '[0-9]{2}(?=%)'
Note the -P option in this case; vanilla grep doesn't seem to support the (?=%) "look-around" part.
The most common way not to capture something is using look-around assertions:
Use it like this
grep -oP '[0-9][0-9](?=%)'
It's worth noting that GNU grep support the -P option to enable Perl compatible regex syntax, however it is not included with OS X. On Linux, it will be available by default. A workaround would be to use ack instead.
But I'd still recommend to use GNU grep on OS X by default. It can be installed on OSX using Homebrew with the command brew grep install
Also, see How to match, but not capture, part of a regex?
You can use sed as an alternative:
sed -rn 's/(^.*)([[:digit:]]{2})(%.*$)/\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 $3+0}'
The value you seek is in the third field, and adding a zero coerces the string to a number, so % is removed.
In bash, I want to get the name of the last folder in a folder path.
For instance, given ../parent/child/, I want "child" as the output.
In a language other than bash, this regex works .*\/(.*)\/$ works.
Here's one of my attempts in bash:
echo "../parent/child/" | sed "s_.*/\(.*?\)/$_\1_p"
This gives me the error:
sed: -e expression #1, char 17: unterminated `s' command
What have I failed to understand?
One problem with your script is that inside the "s_.*/\(.*?\)/$_\1_p" the $_ is interpreted by the shell as a variable name.
You could either replace the double-quotes with single-quotes or escape the $.
Once that's fixed, the .*? may or may not work with your implementation of sed. It will be more robust to write something roughly equivalent that's more widely supported, for example:
sed -e 's_.*/\([^/]*\)/$_\1_'
Note that I dropped the p flag of sed to avoid printing the result twice.
Finally, a much simpler solution will be to use the basedir command.
$ basename ../parent/child/
child
Finally, a native Bash solution is also possible using parameter expansion:
path=../parent/child/
path=${path%/}
path=${path##*/}
You can use cut too
echo '../parent/child/' | cut -d/ -f3
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
Okay, this is an easy one, but I can't figure it out.
Basically I want to extract all links ([^<>]*) from a big html file.
I tried to do this with sed, but I get all kinds of results, just not what I want. I know that my regexp is correct, because I can replace all the links in a file:
sed 's_[^<>]*_TEST_g'
If I run that on something like
<div>A google link</div>
<div>A google link</div>
I get
<div>TEST</div>
<div>TEST</div>
How can I get rid of everything else and just print the matches instead? My preferred end result would be:
A google link
A google link
PS. I know that my regexp is not the most flexible one, but it's enough for my intentions.
Match the whole line, put the interesting part in a group, replace by the content of the group. Use the -n option to suppress non-matching lines, and add the p modifier to print the result of the s command.
sed -n -e 's!^.*\(<[Aa] [^<>]*>.*</[Aa]>\).*$!\1!p'
Note that if there are multiple links on the line, this only prints the last link. You can improve on that, but it goes beyond simple sed usage. The simplest method is to use two steps: first insert a newline before any two links, then extract the links.
sed -n -e 's!</a>!&\n!p' | sed -n -e 's!^.*\(<[Aa] [^<>]*>.*</[Aa]>\).*$!\1!p'
This still doesn't handle HTML comments, <pre>, links that are spread over several lines, etc. When parsing HTML, use an HTML parser.
If you don't mind using perl like sed it can copy with very diverse input:
perl -n -e 's+(<a href=.*?</a>)+ print $1, "\n" +eg;'
Assuming that there is only one hyperlink per line the following may work...
sed -e 's_.*<a href=_<a href=_' -e 's_>.*_>ed <<'EOF'
-e 's_.*<a href=_<a href=_' -e 's_>.*_>_'
This might work for you (GNU sed):
sed '/<a href\>/!d;s//\n&/;s/[^\n]*\n//;:a;$!{/>/!{N;ba}};y/\n/ /;s//&\n/;P;D' file
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.