Use sed to grab a string - regex

I'm using curl to get the html from a site then I just need a specific string which is between 'standards.xml?revision=' and '&amp'. I'm using sed to do this but I can't seem to get the regex right and needed some help.
curl website.com | sed -r 's|.*standards\.xml\?revision=([0-9]+).*|\1|'
The output I'm getting is the full html--any help would be appreciated.

You're almost there. Try using -n option with sed not to print unmatched data and add p modifier to s||| to print replace string
curl website.com | sed -n -r 's|.*standards\.xml\?revision=([0-9]+).*|\1|p'

you can use grep -oP (PCRE option):
grep -oP 'standards\.xml\?revision=\K[0-9]+'
\K resets the matched text hence only later part [0-9]+ is returned.

curl website.com | sed -n '/xml/ {s|.*standards\.xml\?revision=([^&]+).*|\1|p;q;}'
From previous sed [0-9]+ is only if number occur maybe a [^&]+ is more appropriate.
Very good to use the ' and | to avoid problem with \ so I pick it :-)

Related

Why is sed not extracting value?

When I run my regex with sed
echo "abc-def-stg" | sed -e '/(\w*$)/g'
on regexr.com it works with no problems, but when I try to extract the value stg using said it does not work.
Can anyone explain why?
sed is used to replace strings. You are trying to extract.
Use (as John1024 said)
echo "abc-def-stg" | sed '/.*-//'
It will remove all up to and including the last hyphen. Or
echo "abc-def-stg" | grep -oE '[^-]+$'
It will extract all characters other than a hyphen at the end of the string.

What characters do I need to escape with sed to make this regex work

(?<![0-9])0+(?=[0-9]+)
I need to remove unnecessary leading zeros in malformed octettes of IP addresses.
I want to do something like this but it is not working.
cat Qualys-Active-IPs.csv | awk -F';' {'print $1'} | sed 's/(?<![0-9])0+(?\=[0-9]+)//g'
The solution is:
sed -r 's/^0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)$/\1.\2.\3.\4/'
You may try this code:
sed -r 's/^0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)-0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+),...,(.*)$/\9:\1.\2.\3.\4-\5.\6.\7.\8/'

Sed : print all lines after match

I got my research result after using sed :
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | cut -f 1 - | grep "pattern"
But it only shows the part that I cut. How can I print all lines after a match ?
I'm using zcat so I cannot use awk.
Thanks.
Edited :
This is my log file :
[01/09/2015 00:00:47] INFO=54646486432154646 from=steve idfrom=55516654455457 to=jone idto=5552045646464 guid=100021623456461451463 n
um=6 text=hi my number is 0 811 22 1/12 status=new survstatus=new
My aim is to find all users that spam my site with their telephone numbers (using grep "pattern") then print all the lines to get all the information about each spam. The problem is there may be matches in INFO or id, so I use sed to get the text first.
Printing all lines after a match in sed:
$ sed -ne '/pattern/,$ p'
# alternatively, if you don't want to print the match:
$ sed -e '1,/pattern/ d'
Filtering lines when pattern matches between "text=" and "status=" can be done with a simple grep, no need for sed and cut:
$ grep 'text=.*pattern.* status='
You can use awk
awk '/pattern/,EOF'
n.b. don't be fooled: EOF is just an uninitialized variable, and by default 0 (false). So that condition cannot be satisfied until the end of file.
Perhaps this could be combined with all the previous answers using awk as well.
Maybe this is what you actually want? Find lines matching "pattern" and extract the field after text= up through just before status=?
zcat file* | sed -e '/pattern/s/.*text=\(.*\)status=[^/]*/\1/'
You are not revealing what pattern actually is -- if it's a variable, you cannot use single quotes around it.
Notice that \(.*\)status=[^/]* would match up through survstatus=new in your example. That is probably not what you want? There doesn't seem to be a status= followed by a slash anywhere -- you really should explain in more detail what you are actually trying to accomplish.
Your question title says "all line after a match" so perhaps you want everything after text=? Then that's simply
sed 's/.*text=//'
i.e. replace up through text= with nothing, and keep the rest. (I trust you can figure out how to change the surrounding script into zcat file* | sed '/pattern/s/.*text=//' ... oops, maybe my trust failed.)
The seldom used branch command will do this for you. Until you match, use n for next then branch to beginning. After match, use n to skip the matching line, then a loop copying the remaining lines.
cat file | sed -n -e ':start; /pattern/b match;n; b start; :match n; :copy; p; n ; b copy'
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | ***cut -f 1 - | grep "pattern"***
instead change the last 2 segments of your pipeline so that:
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | **awk '$1 ~ "pattern" {print $0}'**

Getting defined substring with help of sed or egrep

Everyone!!
I want to get specific substring from stdout of command.
stdout:
{"response":
{"id":"110200dev1","success":"true","token":"09ad7cc7da1db13334281b84f2a8fa54"},"success":"true"}
I need to get a hex string after token without quotation marks, the length of hex string is 32 letters.I suppose it can be done by sed or egrep. I don't want to use awk here. Because the stdout is being changed very often.
This is an alternate gnu-awk solution when grep -P isn't available:
awk -F: '{gsub(/"/, "")} NF==2&&$1=="token"{print $2}' RS='[{},]' <<< "$string"
09ad7cc7da1db13334281b84f2a8fa54
grep's nature is extracting things:
grep -Po '"token":"\K[^"]+'
-P option interprets the pattern as a Perl regular expression.
-o option shows only the matching part that matches the pattern.
\K throws away everything that it has matched up to that point.
Or an option using sed...
sed 's/.*"token":"\([^"]*\)".*/\1/'
With sed:
your-command | sed 's/.*"token":"\([^"]*\)".*/\1/'
YourStreamOrFile | sed -n 's/.*"token":"\([a-f0-9]\{32\}\)".*/\1/p'
doesn not return a full string if not corresponding

Using sed and regex to capture last part of url

I'm trying to make sed match the last part of a url and output just that. For example:
echo "http://randomurl/suburl/file.mp3" | sed (expression)
should give the output:
file.mp3
So far I've tried sed 's|\([^/]+mp3\)$|\1|g' but it just outputs the whole url. Maybe there's something I'm not seeing here but anyways, help would be much appreciated!
this works:
echo "http://randomurl/suburl/file.mp3" | sed 's#.*/##'
basename is your good friend.
> basename "http://randomurl/suburl/file.mp3"
=> file.mp3
This should do the job:
$ echo "http://randomurl/suburl/file.mp3" | sed -r 's|.*/(.*)$|\1|'
file.mp3
where:
| has been used instead of / to separate the arguments of the s command.
Everything is matched and replaced with whatever if found after the last /.
Edit: You could also use bash parameter substitution capabilities:
$ url="http://randomurl/suburl/file.mp3"
$ echo ${url##*/}
file.mp3
echo 'http://randomurl/suburl/file.mp3' | grep -oP '[^/\n]+$'
Here's another solution using grep.