Currently I'm trying to use sed with regex on Solaris but it doesn't work.
I need to show only lines matching to my regex.
sed -n -E '/^[a-zA-Z0-9]*$|^a_[a-zA-Z0-9]*$/p'
input file:
grtad
a_pitr
_aupa
a__as
baman
12353
ai345
ki_ag
-MXx2
!!!23
+_)#*
I want to show only lines matching to above regex:
grtad
a_pitr
baman
12353
ai345
Is there another way to use alternative? Is it possible in perl?
Thanks for any solutions.
With Perl
perl -ne 'print if /^(a_)?[a-zA-Z0-9]*$/' input.txt
The (a_)? matches a_ one-or-zero times, so optionally. It may or may not be there.
The (a_) also captures the match, what is not needed. So you can use (?:a_)? instead. The ?: makes () only group what is inside (so ? applies to the whole thing), but not remember it.
with grep
$ grep -xiE '(a_)?[a-z0-9]*' ip.txt
grtad
a_pitr
baman
12353
ai345
-x match whole line
-i ignore case
-E extended regex, if not available, use grep -xi '\(a_\)\?[a-z0-9]*'
(a_)? zero or one time match a_
[a-z0-9]* zero or more alphabets or numbers
With sed
sed -nE '/^(a_)?[a-zA-Z0-9]*$/p' ip.txt
or, with GNU sed
sed -nE '/^(a_)?[a-z0-9]*$/Ip' ip.txt
Related
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:]]+)/;'
I have a file with strings similar to this:
abcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'
I have to find current_count and total_count for each line of file. I am trying below command but its not working. Please help.
grep current_count file | sed "s/.*\('current_count': u'\d+'\).*/\1/"
It is outputting the whole line but I want something like this:
'current_count': u'3', 'total_count': u'3'
It's printing the whole line because the pattern in the s command doesn't match, so no substitution happens.
sed regexes don't support \d for digits, or x+ for xx*. GNU sed has a -r option to enable extended-regex support so + will be a meta-character, but \d still doesn't work. GNU sed also allows \+ as a meta-character in basic regex mode, but that's not POSIX standard.
So anyway, this will work:
echo -e "foo\nabcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'" |
sed -nr "s/.*('current_count': u'[0-9]+').*/\1/p"
# output: 'current_count': u'2'
Notice that I skip the grep by using sed -n s///p. I could also have used /current_count/ as an address:
sed -r -e '/current_count/!d' -e "s/.*('current_count': u'[0-9]+').*/\1/"
Or with just grep printing only the matching part of the pattern, instead of the whole line:
grep -E -o "'current_count': u'[[:digit:]]+'
(or egrep instead of grep -E). I forget if grep -o is POSIX-required behaviour.
For me this looks like some sort of serialized Python data. Basically I would try to find out the origin of that data and parse it properly.
However, while being hackish, sed can also being used here:
sed "s/.*current_count': [a-z]'\([0-9]\+\).*/\1/" input.txt
sed "s/.*total_count': [a-z]'\([0-9]\+\).*/\1/" input.txt
Assuming a simple text file:
123.123.123.123
I would like to replace the IP inside of it with 222.222.222.222. I have tried the below but nothing changes, however the same regex seems to work in this Regexr
sed -i '' 's/(\d{1,3}\.){3}\d{1,3}/222.222.222.222/' file.txt
Am I missing something?
Two problems here:
sed doesn't like PCRE digit property \d, use range: [0-9] or POSIX [[:digit:]]
You need to use -r flag for extended regex as well.
This should work:
s='123.123.123.123'
sed -r 's/([0-9]{1,3}\.){3}[0-9]{1,3}/222.222.222.222/' <<< "$s"
222.222.222.222
Better would be to use anchors to avoid matching unexpected input:
sed -r 's/^([0-9]{1,3}\.){3}[0-9]{1,3}$/222.222.222.222/' <<< "$s"
PS: On OSX use -E instead of -r:
sed -E 's/^([0-9]{1,3}\.){3}[0-9]{1,3}$/222.222.222.222/' <<< "$s"
222.222.222.222
You'd better use -r, as indicated by anubhava.
But in case you don't have it, you have to escape every single (, ), { and }. And also, use [0-9] instead of \d:
$ sed 's/\([0-9]\{1,3\}\.\)\{3\}[0-9]\{1,3\}/222.222.222.222/' <<< "123.123.123.123"
222.222.222.222
Let's say I wanted to find all 12-letter words in /usr/share/dict/words that started with c and ended with er. Off the top of my head, a workable pattern could look something like:
grep -E '^c.{9}er$' /usr/share/dict/words
It finds:
cabinetmaker
calcographer
calligrapher
campanologer
campylometer
...
But that .{9} bothers me. It feels too magical, subtracting the total length of all the anchor characters from the number defined in the original constraint.
Is there any way to rewrite this regex so it doesn't require doing this calculation up front, allowing a literal 12 to be used directly in the pattern?
You can use the -x option which selects only matches that exactly match the whole line.
grep -xE '.{12}' | grep 'c.*er'
Ideone Demo
Or use the -P option which clarifies the pattern as a Perl regular expression and use a lookahead assertion.
grep -P '^(?=.{12}$)c.*er$'
Ideone Demo
You can use awk as an alternative and avoid this calculation:
awk -v len=12 'length($1)==len && $1 ~ /^c.*?er$/' file
I don't know grep so well, but some more advanced NFA RegEx implementations provide you with lookaheads and lookbehinds. If you can figure out any means to make those available for you, you could write:
^(?=c).{12}(?<=er)$
Maybe as a perl one-liner like this?
cat /usr/share/dict/words | perl -ne "print if m/^(?=c).{12}(?<=er)$/"
One approach with GNU sed:
$ sed -nr '/^.{12}$/{/^c.*er$/p}' words
With BSD sed (Mac OS) it would be:
$ sed -nE '/^.{12}$/{/^c.*er$/p;}' words
Apologies for the simple question. I don't clean text or use regex often.
I have a large number of text files in which I want to remove every line until my regex finds a match. There's usually about 15 lines of fluff before I find a match. I was hoping for a perl one-liner that would look like this:
perl -p -i -e "s/.*By.unanimous.vote//g" *.txt
But this doesn't work.
Thanks
Solution using the flip-flop operator:
perl -pi -e '$_="" unless /By.unanimous.vote/ .. 1' input-files
Shorter solution that also uses the x=!! pseudo operator:
per -pi -e '$_ x=!! (/By.unanimous.vote/ .. 1)' input-files
Have a try with:
If you want to get rid until the last By.unanimous.vote
perl -00 -pe "s/.*By.unanimous.vote//s" inputfile > outputfile
If you want to get rid until the first By.unanimous.vote
perl -00 -pe "s/.*?By.unanimous.vote//s" inputfile > outputfile
Try something like:
perl -pi -e "$a=1 if !$a && /By\.unanimous\.vote/i; s/.*//s if !$a" *.txt
Should remove the lines before the matched line. If you want to remove the matching line also you can do something like:
perl -pi -e "$a=1 if !$a && s/.*By\.unanimous\.vote.*//is; s/.*//s if !$a" *.txt
Shorter versions:
perl -pi -e "$a++if/By\.unanimous\.vote/i;$a||s/.*//s" *.txt
perl -pi -e "$a++if s/.*By\.unanimous\.vote.*//si;$a||s/.*//s" *.txt
You haven't said whether you want to keep the By.unanimous.vote part, but it sounds to me like you want:
s/[\s\S]*?(?=By\.unanimous\.vote)//
Note the missing g flag and the lazy *? quantifier, because you want to stop matching once you hit that string. This should preserve By.unanimous.vote and everything after it. The [\s\S] matches newlines. In Perl, you can also do this with:
s/.*?(?=By\.unanimous\.vote)//s
Solution using awk
awk '/.*By.unanimous.vote/{a=1} a==1{print}' input > output