#!/bin/bash
TEST="--extra-vars user=jsmith a=abcd --test"
echo $TEST | sed -re "s/(--extra-vars\s)([a-zA-Z0-9\=\s]*)/\1\"\2\"/g"
I am using sed to add literal quotes around the entirety of the second capture group but for some reason its only partially quoting the group. The expected output for this would be --extra-vars "user=jsmith a=abcd" --test but instead I am seeing --extra-vars "user=jsmith" a=abcd --test. I thought using the global sed modifier would help but it hasnt seemed to fix the issue.
Heres a link to the regex101
AFAIK you will need to use [:space:] in character classes:
TEST="--extra-vars user=jsmith a=abcd --test"
echo "$TEST" | sed -re 's/(--extra-vars\s)([a-zA-Z0-9\=[:space:]]*)/\1"\2"/g'
# ^
# Use [:space:]
This will however also include the last space after abcd:
--extra-vars "user=jsmith a=abcd "--test
A way to prevent that is to add (\s-)* to the end of the search and \3 to the replacement:
TEST="--extra-vars user=jsmith a=abcd --test"
echo "$TEST" | sed -re 's/(--extra-vars\s)([a-zA-Z0-9\=[:space:]]*)(\s-)*/\1"\2"\3/g'
Which gives:
--extra-vars "user=jsmith a=abcd" --test
You can also consider using something like Perl which have PCRE:
echo "$TEST" | perl -pe 's/(?<=--extra-vars\s).*?(?=\s-|$)/"$&"/'
To get the output you posted from the input you posted is just:
TEST="--extra-vars user=jsmith a=abcd --test"
$ echo "$TEST" | sed 's/ \(.*\) / "\1" /'
--extra-vars "user=jsmith a=abcd" --test
If that's not adequate then edit your question to show some more truly representative sample input and expected output.
Notice that if you use [:space:] or an actual space character (i.e. ([a-zA-Z0-9\= ]*) for your second group, it will do what you want, except it'll match that space after abcd:
--extra-vars "user=jsmith a=abcd "--test
So use one of the two, but you'll have to adjust your regex to do what you want. It's difficult for us to give you a regular expression because we aren't sure what kind of string you're expecting, but this might be able to work:
echo "$TEST" | sed -re "s!(--extra-vars\s)([a-zA-Z0-9\=]*\s)([a-zA-Z0-9\=]*)!\1\"\2\3\"!g"
Related
I have working example of substitution in online regex tester https://regex101.com/r/3FKdLL/1 and I want to use it as a substitution in sed editor.
echo "repo-2019-12-31-14-30-11.gz" | sed -r 's/^([\w-]+)-\d{4}-\d{2}-\d{2}-\d{2}-\d{2}-\d{2}.gz$.*/\1/p'
It always prints whole string: repo-2019-12-31-14-30-11.gz, but not matched group [\w-]+.
I expect to get only text from group which is repo string in this example.
Try this:
echo "repo-2019-12-31-14-30-11.gz" |
sed -rn 's/^([A-Za-z]+)-[[:alnum:]]{4}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}.gz.*$/\1/p'
Explanations:
\w will work (not [\w] wich matches either backslash or w), but you should use [[:alnum:]] which is POSIX
For sed, \d isn't a regex class, but an escaped character representing a non-printable character
Add -n to mute sed, with /p to explicitly print matched lines
Additionaly, you could refactor your regex by removing duplication:
echo "repo-2019-12-31-14-30-11.gz" |
sed -rn 's/^([[:alnum:]]+)-[[:digit:]]{4}(-[[:digit:]]{2}){5}.gz.*$/\1/p'
Looks like a job for GNU grep :
echo "repo-2019-12-31-14-30-11.gz" | grep -oP '^\K[[:alpha:]-]+'
Displays :
repo-
On this example :
echo "repo-repo-2019-12-31-14-30-11.gz" | grep -oP '^\K[[:alpha:]-]+'
Displays :
repo-repo-
Which I think is what you want because you tried with [\w-]+ on your regex.
If I'm wrong, just replace the grep command with : grep -oP '^\K\w+'
I can't seem to get a substring correctly.
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g')
That still returns bugfix/US3280841-something-duh.
If I try an use perl instead:
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9]|[A-Z0-9])+/; print $1');
That outputs nothing.
What am I doing wrong?
Using bash parameter expansion only:
$: # don't use caps; see below.
$: declare branch="bugfix/US3280841-something-duh"
$: tmp="${branch##*/}"
$: echo "$tmp"
US3280841-something-duh
$: trimmed="${tmp%%-*}"
$: echo "$trimmed"
US3280841
Which means:
$: tmp="${branch_name##*/}"
$: trimmed="${tmp%%-*}"
does the job in two steps without spawning extra processes.
In sed,
$: sed -E 's#^.*/([^/-]+)-.*$#\1#' <<< "$branch"
This says "after any or no characters followed by a slash, remember one or more that are not slashes or dashes, followed by a not-remembered dash and then any or no characters, then replace the whole input with the remembered part."
Your original pattern was
's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g'
This says "remember any number of anything followed by a slash, then a lowercase letter or a digit, then a pipe character (because those only work with -E), then a capital letter or digit, then a literal plus sign, and then replace it all with what you remembered."
GNU's manual is your friend. I look stuff up all the time to make sure I'm doing it right. Sometimes it still takes me a few tries, lol.
An aside - try not to use all-capital variable names. That is a convention that indicates it's special to the OS, like RANDOM or IFS.
You may use this sed:
sed -E 's~^.*/|-.*$~~g' <<< "$BRANCH_NAME"
US3280841
Ot this awk:
awk -F '[/-]' '{print $2}' <<< "$BRANCH_NAME"
US3280841
sed 's:[^/]*/\([^-]*\)-.*:\1:'<<<"bugfix/US3280841-something-duh"
Perl version just has + in wrong place. It should be inside the capture brackets:
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9A-Z]+)/; print $1');
Just use a ^ before A-Z0-9
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[^A-Z0-9]\+/\1/g')
in your sed case.
Alternatively and briefly, you can use
TRIMMED=$(echo $BRANCH_NAME | sed "s/[a-z\/\-]//g" )
too.
type on shell terminal
$ BRANCH_NAME="bugfix/US3280841-something-duh"
$ echo $BRANCH_NAME| perl -pe 's/.*\/(\w\w[0-9]+).+/\1/'
use s (substitute) command instead of m (match)
perl is a superset of sed so it'd be identical 'sed -E' instead of 'perl -pe'
Another variant using Perl Regular Expression Character Classes (see perldoc perlrecharclass).
echo $BRANCH_NAME | perl -nE 'say m/^.*\/([[:alnum:]]+)/;'
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.
So I have a bash script which is working perfectly except for one issue with sed.
full=$(echo $full | sed -e 's/\b'$first'\b/ /' -e 's/ / /g')
This would work great except there are instances where the variable $first is preceeded immediately by a period, not a blank space. In those instances, I do not want the variable removed.
Example:
full="apple.orange orange.banana apple.banana banana";first="banana"
full=$(echo $full | sed -e 's/\b'$first'\b/ /' -e 's/ / /g')
echo $first $full;
I want to only remove the whole word banana, and not make any change to orange.banana or apple.banana, so how can I get sed to ignore the dot as a delimiter?
You want "banana" that is preceded by beginning-of-string or a space, and followed by a space or end-of-string
$ sed -r 's/(^|[[:blank:]])'"$first"'([[:blank:]]|$)/ /g' <<< "$full"
apple.orange orange.banana apple.banana
Note the use of -r option (for bsd sed, use -E) that enables extended regular expressions -- allow us to omit a lot of backslashes.
If I have a string that contains this somewhere (Foo could be anything):
<tag>Foo</tag>
How would I, using SED and RegEx, replace it with this:
[tag]Foo[/tag]
My failed attempt:
echo "<tag>Foo</tag>" | sed "s/<tag>\(.*\)<\\/tag>/[tag]\1[\\/tag]"
Your regex is missing the terminating /
$ echo "<tag>Foo</tag>" | sed "s/<tag>\(.*\)<\\/tag>/[tag]\1[\\/tag]/"
[tag]Foo[/tag]
With this you can replace all types of tags and don't have to be tag specific.
$echo "<tag>Foo</tag>" | sed "s/[^<]*<\([^>]*\)>\([^<]*\)<\([^>]*\)>/[\1]\2[\3]/"
hope this helps.