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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']
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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.
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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)
}
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I want to match the following with multiple capture groups:
Definition 1
: This is the definition text that described the term. Can have markdown formatting and
multiple lines.
Definition 2
: This is the definition text with **markdown**.`code`
I also want to replace it with the following text (HTML definition list):
<dl>
<dt>Definition 1</dt>
<dd>This is the definition text that described the term. Can have markdown formatting and
multiple lines.</dd>
<dt>Definition 1</dt>
<dd>This is the definition text with **markdown**.`code`</dd>
</dl>
You could do this in two steps:
1. Insert the dt and dd tags
Perform a search with:
(.*)\R: ((?:.+(?:\R|$))*?)(?=\R|.*\R:|$)/g
and substitute by:
<dt>$1</dt>\n<dd>$2</dd>\n
See regex tester.
2. Add the dl tags
Use the result of the previous substitution and perform the following search:
/(<dt>.*?<\/dd>(?!\s*<dt>))/gs
and substitute by:
<dl>\n$1\n</dl>
See regex tester.
Remarks
If the \R escape is not supported, use \n instead.
The back-references $1, $2 might need to be changed to \1, \2 depending on your regex engine.
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I cannot figure out how to match only on groups that contain a certain word ('test' for example below). It is a big text file and the groups start with a line 'Group x' and include text with an empty line separation to the next group. I think I need to use lookaheads and lookbehinds but don't know how. I can use vb.net for this but trying to test out different expressions in the regex testers and can't get anywhere.
Group 1
adfdf
dd test ddfdf
dfdfadf
Group 2
ddfadfa
Group 3
add test
adfdff
Group 4
adfdf
Expected 2 matches:
Group 1
adfdf
dd test ddfdf
dfdfadf
Group 3
add test
adfdff
Start your pattern with ^Group \d+$ and end with (?:^$|\Z). In the middle match test but not preceeded by an empty line $(?:.(?!^$) (see Regular expression to match a line that doesn't contain a word? for details on how the latter works). Don't forget the m and s modifiers:
^Group \d+$(?:.(?!^$))*?test.*?(?:^$|\Z)
Demo: https://regex101.com/r/kM9qB3/2
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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