Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 6 years ago.
Improve this question
I have a groovy value like the following:
"hostname blah blah blah"
When that occurs, I want to keep the hostname, but delete everything after it including the first space.
I'm having difficulty with the regex. Any ideas?
Thanks!
To extract the hostname and/or a part of a string by a regular expression can be done in a straightforward way.
If the values are separated by whitespace then can just split the string by a space character and use the first part. Split() method takes in a regular expression as its first argument.
String s = "hostname blah blah blah"
String[] parts = s.split(' ', 2)
if (parts) {
println parts[0]
s = parts[0] // if want to replace the variale s with just the hostname
}
Here is an alternate example using a regular expression to pull out the hostname part of the string value. The first part of the regular expression (\S+) is looking for a sequence of non-whitespace characters. For a set of possible hostnames, a stricter expression could be something like (\w+(\.\w+)*).
import java.util.regex.Matcher
if (s =~ /^(\S+)\s/) {
println Matcher.lastMatcher.group(1)
}
Related
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 12 hours ago.
The community is reviewing whether to reopen this question as of 9 hours ago.
Improve this question
I want to extract the words in given string which are start with '#' in notepad++. Can any one provide your suggestions/solutions?
Here is the string:
#Data is a convenient shortcut annotation that bundles the features of #ToString, #EqualsAndHashCode, #Getter / #Setter and #RequiredArgsConstructor together: In other words
Using Notepad++:
Refer to this answer: Regular expressions in notepad++ (Search and Replace). Open up "Find and Replace" menu. Set search mode to "Regular expression". Then search for your regular expression #\w+ in the "Find what" field.
Another option is to use Python:
Try the re Python module to extract the words from the variable using the same regular expression pattern. The regular expression pattern #\w+ matches any word starting with # followed by one or more word characters.
import re
string = "#Data is a convenient shortcut annotation that bundles the features of #ToString, #EqualsAndHashCode, #Getter / #Setter and #RequiredArgsConstructor together: In other words"
matches = re.findall(r'#\w+', string)
print(matches)
Outputs:
['#Data', '#ToString', '#EqualsAndHashCode', '#Getter', '#Setter', '#RequiredArgsConstructor']
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 4 years ago.
Improve this question
Please can somebody help me, I`m new to regex and have no idea how to do this!.
I`m trying to extract from a list which looks like this...
Joe-Age23-46737-251.aspx
Tim-Age18-46909-451.aspx
Roger-Age41-59768-251.aspx
What I want is this...
46737-251.aspx
46909-451.aspx
59768-251.aspx
so basically anything after the second to last hyphen.
Cheers
Let's translate "everything after the second-to-last hyphen" into regex:
(?<=-)[^-]*-[^-]*$
Explanation:
(?<=-) # Assert starting position right after a hyphen
[^-]* # Match zero or more characters except hyphens
- # Match a single hyphen
[^-]* # see above
$ # until end of string.
Test it live on regex101.com.
Step1 : Split the string on the basis of hyphen(-) . You will get array of strings.
Step2 : extract the second , fifth and eighth
and so on( incremented by 3 ).
Step3 : concatinate all the strings formed in step2.
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 4 years ago.
Improve this question
I am trying to find a pattern to remove % from the text. This could have been easily achieved through ESCAPE or replace but I am restricted to only modify the contents of $pattern .
$text = "something is 5% and 10% value"
$pattern = "[^!%]*" // only this can be modified.
([Regex]::Match($text,$pattern)).value
Output should be :
something is 5 and 10 value
From the documentation:
Remarks
The Match(String, String) method returns the first substring that matches a regular expression pattern in an input string.
If you can only modify the pattern but not the rest of the code it's not possible to achieve what you want.
You need somthing like
$pattern = '[!%]'
[regex]::Replace($text, $pattern, '')
or
$pattern = '[!%]'
$text -replace $pattern
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 9 years ago.
Improve this question
I want to be able to match all of the following strings to my regex below. It doesnt seem to be working. Any suggestions?
Strings to compare :
5878ce43aa3f1e1d713427d118115310 -1 Script Kiddie <perm>
f939f88b50fa5f0099b6751e7be27761 -1 Hacking <perm>
468f6634c5a9b00b5b3872dd6437143f 1356474103 Being Annoying <7day>
This is my perl code. It isnt working at the moment. Any suggestions?
my $bn_re = q{(.+?) (\d+) (.+?)};
If the first two fields are always without whitespace in them, you can use split to great effect, using the LIMIT option to only get three fields:
my ($str, $num, $other) = split ' ', $_, 3;
That is, assuming you read the file something like this:
while (<>) {
... # your code here
}
Also, this:
my $bn_re = q{(.+?) (\d+) (.+?)};
is not a regex. You may be confusing q() with qr(). You may also be confusing the functionality of
$str =~ $bn_re;
Which will automagically include the regex in a match operator m//. But you should use qr(). The q() operator does what the single quote does.
Also, you should be aware that .+? will match a single char if you allow it. As it does at the end of your "regex". At the end of your string, either do
... (.+)/ # matching greedily
... (.+?)$/ # using anchor to end of string
$bn_re =~ /[0-9a-z]+?\s[-0-9]+\s[\w\s]+?[<>a-z0-9]+?/i
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
I have a big list of simple expressions (2Mb file). For example:
11.*;112.*;113.*;12.*;123.*
I need to remove unnecessary expressions and come up with this:
11.*;12.*
bash version would be appreciated. thanks in advance
Here is something in Perl that would work, provided the only wildcards in your pattern are of the form .*:
#!/usr/bin/perl
use strict;
use warnings;
my %terms;
{
local $/;
%terms = map {$_ => 1} split /;|\n/, <>;
}
foreach my $k1 (keys %terms)
{
foreach my $k2 (keys %terms)
{
if ($k1 ne $k2 and $k1 =~ /^$k2$/)
{
delete $terms{$k1};
last;
}
}
}
print join ';', keys %terms;
It accepts your file as a command line argument.
This works by comparing keys to each other. In each comparison, one key is treated as a string and the other key is evaluated as a regex. This takes advantage of the fact the .* matches anything--including the literal characters .*. Thus an expression that matches the literal string of another pattern will also match all strings that pattern would match.
It will work even if there are multiple .* terms in a single pattern. For example, it correctly determines that 1.*1.* matches everything that 11.* matches, deleting the latter.
However, this is kind of a hacky simplification and will not work if you introduce other regex patterns. There is no easy solution to this problem in general, because you would have to parse all patterns and figure out what each would match.