I just started using sed from doing regex. I wanted to extract XXXXXX from *****/XXXXXX> so I was following
sed -n "/^/*/(\S*\).>$/p"
If I do so I get following error
sed: 1: "/^//(\S).>$/p": invalid command code *
I am not sure what am I missing here.
Try:
$ echo '*****/XXXXXX>' | sed 's|.*/||; s|>.*||'
XXXXXX
The substitute command s|.*/|| removes everything up to the last / in the string. The substitute command s|>.*|| removes everything from the first > in the string that remains to the end of the line.
Or:
$ echo '*****/XXXXXX>' | sed -E 's|.*/(.*)>|\1|'
XXXXXX
The substitute command s|.*/(.*)>|\1| captures whatever is between the last / and the last > and saves it in group 1. That is then replaced with group 1, \1.
In my opinion awk performs better this task. Using -F you can use multiple delimiters such as "/" and ">":
echo "*****/XXXXXX>" | awk -F'/|>' '{print $1}'
Of course you could use sed, but it's more complicated to understand. First I'm removing the first part (delimited by "/") and after the second one (delimited by ">"):
echo "*****/XXXXXX>" | sed -e s/.*[/]// -e s/\>//
Both will bring the expected result: XXXXXX.
with grep if you have pcre option
$ echo '*****/XXXXXX>' | grep -oP '/\K[^>]+'
XXXXXX
/\K positive lookbehind / - not part of output
[^>]+ characters other than >
echo '*****/XXXXXX>' |sed 's/^.*\/\|>$//g'
XXXXXX
Start from start of the line, then proceed till lask / ALSO find > followed by EOL , if any of these found then replace it with blank.
Related
I have the string like this feature/test-111-test-test.
I need to extract string till the second dash and change forward slash to dash as well.
I have to do it in Makefile using shell syntax and there for me doesn't work some regular expression which can help or this case
Finally I have to get smth like this:
input - feature/test-111-test-test
output - feature-test-111- or at least feature-test-111
feature/test-111-test-test | grep -oP '\A(?:[^-]++-??){2}' | sed -e 's/\//-/g')
But grep -oP doesn't work in my case. This regexp doesn't work as well - (.*?-.*?)-.*.
Another sed solution using a capture group and regex/pattern iteration (same thing Socowi used):
$ s='feature/test-111-test-test'
$ sed -E 's/\//-/;s/^(([^-]*-){3}).*$/\1/' <<< "${s}"
feature-test-111-
Where:
-E - enable extended regex support
s/\//-/ - replace / with -
s/^....*$/ - match start and end of input line
(([^-]-){3}) - capture group #1 that consists of 3 sets of anything not - followed by -
\1 - print just the capture group #1 (this will discard everything else on the line that's not part of the capture group)
To store the result in a variable:
$ url=$(sed -E 's/\//-/;s/^(([^-]*-){3}).*$/\1/' <<< "${s}")
$ echo $url
feature-test-111-
You can use awk keeping in mind that in Makefile the $ char in awk command must be doubled:
url=$(shell echo 'feature/test-111-test-test' | awk -F'-' '{gsub(/\//, "-", $$1);print $$1"-"$$2"-"}')
echo "$url"
# => feature-test-111-
See the online demo. Here, -F'-' sets the field delimiter as -, gsub(/\//, "-", $1) replaces / with - in Field 1 and print $1"-"$2"-" prints the value of --separated Field 1 and 2.
Or, with a regex as a field delimiter:
url=$(shell echo 'feature/test-111-test-test' | awk -F'[-/]' '{print $$1"-"$$2"-"$$3"-"}')
echo "$url"
# => feature-test-111-
The -F'[-/]' option sets the field separator to - and /.
The '{print $1"-"$2"-"$3"-"}' part prints the first, second and third value with a separating hyphen.
See the online demo.
To get the nth occurrence of a character C you don't need fancy perl regexes. Instead, build a regex of the form "(anything that isn't C, then C) for n times":
grep -Eo '([^-]*-){2}' | tr / -
With sed and cut
echo feature/test-111-test-test| cut -d'-' -f-2 |sed 's/\//-/'
Output
feature-test-111
echo feature/test-111-test-test| cut -d'-' -f-2 |sed 's/\//-/;s/$/-/'
Output
feature-test-111-
You can use the simple BRE regex form of not something then that something which is [^-]*- to get all characters other than - up to a -.
This works:
echo 'feature/test-111-test-test' | sed -nE 's/^([^/]*)\/([^-]*-[^-]*-).*/\1-\2/p'
feature-test-111-
Another idea using parameter expansions/substitutions:
s='feature/test-111-test-test'
tail="${s//\//-}" # replace '/' with '-'
# split first field from rest of fields ('-' delimited); do this 3x times
head="${tail%%-*}" # pull first field
tail="${tail#*-}" # drop first field
head="${head}-${tail%%-*}" # pull first field; append to previous field
tail="${tail#*-}" # drop first field
head="${head}-${tail%%-*}-" # pull first field; append to previous fields; add trailing '-'
$ echo "${head}"
feature-test-111-
A short sed solution, without extended regular expressions:
sed 's|\(.*\)/\([^-]*-[^-]*\).*|\1-\2|'
I have done lots of searching on SO and google.
I am using a regex tester like Regex 101.
The test string is [../../dvl/namespaces/my-namespace-1/notterminating]
The regex I am using is .*\/([^\/]*)[\[\]']+.
What I am trying to do (returns empty):
$ param="[../../dvl/namespaces/my-namespace-1/notterminating]"
$ echo $param | grep ".*\/([^\/]*)[\[\]']+"
I have tried different options with grep by adding different flags like -o, -e, etc. I have tried "${param}.*\/([^\/]*)[\[\]']+". However, I have been unsuccessful at getting bash to recognize my regex.
You may use sed:
sed -n "s#.*/\([^]['/]*\).*#\1#p" <<< "$param"
See an online demo
Details
.* - matches 0+ chars
/ - / char
\([^]['/]*\) - Group 1: any 0+ chars other than ], [, ' and /
.* - any 0+ chars.
This way, the whole string is replaced with the Group 1 value, and only that value remains as -n suppresses default line output and p prints the result.
Without using any external tool you can do this in pure bash with IFS:
IFS='/\[\]' read -ra arr <<< "$param" && echo "${arr[-1]}"
notterminating
Otherwise you may use this simple sed:
sed 's~.*/~~' <<< "${param//[]]}"
notterminating
Or by using awk:
awk -F '[][/]' '{print $(NF-1)}' <<< "$param"
notterminating
Using sed:
echo "$param" |sed -r -n 's,.*\/(.*)\],\1,p'
output:
notterminating
I'm trying to get substrings using sed with regexp. I want to get the first and second "fields" delimited by ":".
To get the first field I used the following command, but don't know how to get the second field.
Command used to get the first field:
sed -r -n '1,2 s/([^:]+).*/\1/p' /etc/passwd
Input file (example):
root:x:0:0:root:/root:/bin/bash
daemon:x:1:1:daemon:/usr/sbin:/usr/sbin/nologin
Command's result:
root
daemon
But I tried do get the first ("root") and second ("x") fields (examples based on the file's first line only), but I did't succedded.
I tried:
sed -r -n '1,2 s/([^:]+).*([^:]+).*/\1 \2/p' /etc/passwd
Command's result:
root h
daemon n
Desired result:
root x
daemon x
sed uses greedy match. In
sed -r -n '1,2 s/([^:]+).*([^:]+).*/\1 \2/p' /etc/passwd
^^
.* matches as many characters as possible. You need
sed -r -n '1,2 s/([^:]+):([^:]+).*/\1 \2/p' /etc/passwd
^
Demo: http://ideone.com/wjL7Za.
By the way, a simpler way to do this is using cut:
cut -d ":" -f 1,2 --output-delimiter=' ' /etc/passwd
Demo: http://ideone.com/stJdSy.
Another expression that would return the desire result would be:
([a-z]+):([a-z]+).*
RegEx Demo
sed -r -n '1,2 s/([^:]+):([^:]+).*/\1 \2/p'
Sed Demo
I am running a linux bash script. From stout lines like: /gpx/trk/name=MyTrack1, I want to keep only the end of line after =.
I am struggling to understand why the following sed command is not working as I expect:
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
(I also tried)
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*\=//"
The return is always /gpx/trk/name=MyTrack1 and not MyTrack1
An even simpler way if this is the only structure you are concerned about:
echo "/gpx/trk/name=MyTrack1" | cut -d = -f 2
Simply try:
echo "/gpx/trk/name=MyTrack1" | sed 's/.*=//'
Solution 2nd: With another sed.
echo "/gpx/trk/name=MyTrack1" | sed 's/\(.*=\)\(.*\)/\2/'
Explanation: As per OP's request adding explanation for this code here:
s: Means telling sed to do substitution operation.
\(.*=\): Creating first place in memory to keep this regex's value which tells sed to keep everything in 1st place of memory from starting to till = so text /gpx/trk/name= will be in 1 place.
\(.*\): Creating 2nd place in memory for sed telling it to keep everything now(after the match of 1st one, so this will start after =) and have value in it as MyTrack1
/\2/: Now telling sed to substitute complete line with only 2nd memory place holder which is MyTrack1
Solution 3rd: Or with awk considering that your Input_file is same as shown samples.
echo "/gpx/trk/name=MyTrack1" | awk -F'=' '{print $2}'
Solution 4th: With awk's match.
echo "/gpx/trk/name=MyTrack1" | awk 'match($0,/=.*$/){print substr($0,RSTART+1,RLENGTH-1)}'
$ echo "/gpx/trk/name=MyTrack1" | sed -e "s/^.*=//"
MyTrack1
The regular expression ^.*= matches anything up to and including the last = in the string.
Your regular expression ^*= would match the literal string *= at the start of a string, e.g.
$ echo "*=/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
/gpx/trk/name=MyTrack1
The * character in a regular expression usually modifies the immediately previous expression so that zero or more of it may be matched. When * occurs at the start of an expression on the other hand, it matches the character *.
Not to take you off the sed track, but this is easy with Bash alone:
$ echo "$s"
/gpx/trk/name=MyTrack1
$ echo "${s##*=}"
MyTrack1
The ##*= pattern removes the maximal pattern from the beginning of the string to the last =:
$ s="1=2=3=the rest"
$ echo "${s##*=}"
the rest
The equivalent in sed would be:
$ echo "$s" | sed -E 's/^.*=(.*)/\1/'
the rest
Where #*= would remove the minimal pattern:
$ echo "${s#*=}"
2=3=the rest
And in sed:
$ echo "$s" | sed -E 's/^[^=]*=(.*)/\1/'
2=3=the rest
Note the difference in * in Bash string functions vs a sed regex:
The * in Bash (in this context) is glob like - itself means 'any character'
The * in a regex refers to the previous pattern and for 'any character' you need .*
Bash has extensive string manipulation functions. You can read about Bash string patterns in BashFAQ.
Using Sed I tried but it did not worked out.
Basically, I have a string say:-
Input:-
'http://www.google.com/photos'
Output required:-
http://www.google.com
I tried using sed but escaping ' is not possible.
what i did was:-
sed 's/\'//' | sed 's/photos//'
sed for photos worked but for ' it didn't.
Please suggest what can be the solution.
Escaping ' in sed is possible via a workaround:
sed 's/'"'"'//g'
# |^^^+--- bash string with the single quote inside
# | '--- return to sed string
# '------- leave sed string and go to bash
But for this job you should use tr:
tr -d "'"
Perl Replacements have a syntax identical to sed, works better than sed, is installed almost in every system by default and works for all machines the same way (portability):
$ echo "'http://www.google.com/photos'" |perl -pe "s#\'##g;s#(.*//.*/)(.*$)#\1#g"
http://www.google.com/
Mind that this solution will keep only the domain name with http in front, discarding all words following http://www.google.com/
If you want to do it with sed , you can use sed "s/'//g" as advised by Wiktor Stribiżew in comments.
PS: I sometimes refer to special chars with their ascii hex code of the special char as advised by man ascii, which is \x27 for '
So for sed you can do it:
$ echo "'http://www.google.com/photos'" |sed -r "s#'##g; s#(.*//.*/)(.*$)#\1#g;"
http://www.google.com/
# sed "s#\x27##g' will also remove the single quote using hex ascii code.
$ echo "'http://www.google.com/photos'" |sed -r "s#'##g; s#(.*//.*)(/.*$)#\1#g;"
http://www.google.com #Without the last slash
If your string is stored in a variable, you can achieve above operations with pure bash, without the need of external tools like sed or perl like this:
$ a="'http://www.google.com/photos'" && a="${a:1:-1}" && echo "$a"
http://www.google.com/photos
# This removes 1st and last char of the variable , whatever this char is.
$ a="'http://www.google.com/photos'" && a="${a:1:-1}" && echo "${a%/*}"
http://www.google.com
#This deletes every char from the end of the string up to the first found slash /.
#If you need the last slash you can just add it to the echo manually like echo "${a%/*}/" -->http://www.google.com/
It's unclear if the ' are actually around your string, although this should take care it:
str="'http://www.google.com/photos'"
echo "$str" | sed s/\'//g | sed 's/\/photos//g'
Combined:
echo "$str" | sed -e "s/'//g" -e 's/\/photos//g'
Using tr:
echo "$str" | sed -e "s/\/photos//g" | tr -d \'
Result:
http://www.google.com
If the single quotes are not around your string it should work regardless.