My bad, I have updated the question-its using Powershell
my file contains 1000s of lines like below:
<dependency org="${abcd}" name="some-random-name" rev="100.100" conf="compile;runtime"/>
I would like to get only the output like:
name="some-random-name"
how can i achieve this. please help
This probably will solve your issue:
cat <file> | grep -oP 'name="[\w-]*"'
Explaining:
grep is the tool that print lines matching a pattern
-o option will print only the matching parts
-P option will use Perl-style regex in order to allow the \w metacharacter.
[\w-]* will match any string containing only 'word' characters or dash with size >= 0
Related
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20
I have data as this
tatusx2.atc?beginnum=0;8pctgRB Mwdf fgEio"text1"text4"text
tatqsx3.atc?beginnum=1;8pctgRBwsaNezxio"text2
tatssx4.atc?beginnum=2;8pctgsvMALNejkio"data2
tatksx4.atc?beginnum=1;8pctgxdfALNebfio"text3
tatzsx5.atc?beginnum=3;8pwerRBMALNetior"datac
How to get only data between ; and "
I have tried grep -oP ';.*?"' file and got output :
;8pctgRBMwdffgEio"
;8pctgRBwsaNezxio"
;8pctgsvMALNejkio"
;8pctgxdfALNebfio"
;8pwerRBMALNetior"
But my desired output is:
8pctgRB Mwdf fgEio
8pctgRBwsaNezxio
8pctgsvMALNejkio
8pctgxdfALNebfio
8pwerRBMALNetior
You need to use lookahead and lookbehind regex expressions
grep -oP '(?<=;)\w*(?=")'
I consider you play around regexr to learn more about regular expressions. Checkout their cheatsheet.
A much more readable way to write the expression you need is:
grep -oP '(?<=;).*(?=")' file
and will get you the desired result. PERL regexes are apparently experimental but certain patterns work without issues.
The following options are being used:
-o --only-matching to the print only the matched parts of a matching line
-P --perl-regexp
Using ?=; will get you the string beginning with ; but using the > you are able to start at the index after. Similarly the end string tag is specified.
Here is suggested additional reading.
I want to do something like
egrep -o '(mon|tues)[1-3]?[0-9].*(mon|tues)[1-3]?[0-9]'
And only get what isn't found by the (mon|tues)[1-3]?[0-9]
With this as input
mon19hellotues20
mon19world
hellomon19
tues8worldtues22
I want
mon19tues20
tues8tues22
As output
sed is better tool for this to print certain matched txt in output:
sed -nE 's/(mon|tues)([1-3]{0,1}[0-9]).*(mon|tues)([1-3]{0,1}[0-9])/\1\2\3\4/p' file
mon19tues20
tues8tues22
I want to match the below string using a regular expression in grep command.
File name is test.txt,
Unknown Unknown
Jessica Patiño
Althea Dubravsky 45622
Monique Outlaw 49473
April Zwearcan 45758
Tania Horne 45467
I want to match the lines containing special characters alone from the above list of lines; the line which I exactly need is 'Jessica Patiño', which contains a non-ASCII character.
I used,
grep '[^0-9a-zA-Z]' test.txt
But it returns all lines.
The following command should return the lines you want:
grep -v '^[0-9a-zA-Z ]*$' test.txt
Explanation
[0-9a-zA-Z ] matches a space or any alphanumeric character.
Adding the asterisk matches any string containing only these characters.
Prepending the pattern with ^ and appending it with $ anchors the string to the beginning and end of line so that the pattern matches only the lines which contain only the desired characters.
Finally, the -v or --invert-match option to grep inverts the sense of matching, i.e., select non-matching lines.
The provided answers should work for the example text given. However, you're likely to come across people with hyphens or apostrophes in their names, etc. To search for all non-ASCII characters, this should do the trick:
grep -P "[\x00-\x1F\x7F-\xFF]" test.txt
-P enables "Perl" mode and allows use of character code searches. \x00-\x1F are control characters, and \x7F-\xFF is everything above 126.
I would use:
grep [^0-9a-zA-Z\s]+ test.txt
live example
Or, even better:
grep -i "[^\da-z\s]" test.txt
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20