I've got (To) [a-z]+ as regular expression and I've got sentence: To kot dziki pies.
And If I compile it I will retrieve To kot.
So what can I do to retrieve only word after (only kot) "To" instead of "To kot"?
^To (\w+)$ should do the trick. \w is shorthand for any word character, eg. a-z in English; other characters in other languages. If you put parens around To as in your example, it will create a matched group, which means the match for [a-z]+ will be in the second group, and To will be in the first group.
I can really recommend using an interactive tool for testing and developing regular expression, such as Expresso.
use groups inside the regular expression -
"To ([a-z]+)" and then retrieve group 1's value, it will contain "kot" supplied your example string.
Or you could use lookbehinds:
(?<=To )[a-z]+
Then the To does not become part of the captured expression.
Related
Practicing some regex.
Trying to only get Regular, Expressions, and abbreviated
from the below data
Regular Expressions, abbreviated as Regex or Regexp, are a string of characters created within the framework of Regex syntax rules.
With (\w+\S?), I get all words including a nonwhitespace character if present.
How would I get just Regular, Expressions, , and abbreviated ?
Edit:
To clarify, I'm looking for
Regex Expressions, abbreviated separately without spaces
not Regex Expressions, abbreviated (spaces included here)
Regex can't "select". It can only match and capture.
This captures the first 3 words (including optional trailing comma) as groups 1, 2 and 3:
^(\w+,?)\s+(\w+,?)\s+(\w+,?)
See live demo.
as #Bohemian has pointed out, in regex you cannot select but rather capture. If the Regex implementation that you use supports it, then captured group will be returned as part of the match. For example in JS this will happen giving you the results separated.
Capturing groups are created by grouping in parenthesis the part of the match that you want to take out
To match those three specific words the regex would be the following
/(Regular) (Expressions), (abbreviated)/
Note that the words you care about are inside the parenthesis, while the parts of the string you don't want (like spaces and comas) are outside the string
You would use it like this (javascript code)
const string = "Regular Expressions, abbreviated as Regex or Regexp, are a string of characters created within the framework of Regex syntax rules."
const regex = /(Regular) (Expressions), (abbreviated)/;
string.match(regex); // returns [ "Regular Expressions, abbreviated", "Regular", "Expressions", "abbreviated" ]
Note that in the result the first element is the whole match, and the 2nd, 3rd and 4rh element are your capture groups that you can use as if you had selected them from the string
To match any three words separated by space or coma you could use
/(\w+),?\s?(\w+),?\s?(\w+),?\s?/
\w represents a char
\s represents a space
? indicates that there might be 0 or 1 ocurrence of what is previews
and finally the parenthesis group the word and leave out everything else the same as the example above
You would use it like this (javascript code)
const string = "Regular Expressions, abbreviated as Regex or Regexp, are a string of characters created within the framework of Regex syntax rules."
const regex = /(\w+),?\s?(\w+),?\s?(\w+),?\s?/;
string.match(regex); // returns [ "Regular Expressions, abbreviated", "Regular", "Expressions", "abbreviated" ]
I have the below regular expression which retrieves me all characters begins with
(state%3)((?:(?!#).)*)
I want to ignore the state%3. I have tried all kinds of lookback but nothing seems to work
Here is the full text that I need to match agains
"state%3DnGl%252BlPm8CkHfYd2PpBq7W0H2z6xgUeICgB7KFmGmGG8cTSQTf%252B9cYCfFSsT5YSPTITdbaLAlJoQ22%252FCXRAu3ROqTQYzpPfGYxKmRZ7iIqwx3g0GLpVkaXq5FL3Js5FcTGpncQx7TA9w1A6HsSyxxcktfwX8QSzhqJQj5lntOolrPoIqpa4l2C%252BbhCWuAOY18BwVynMv8%252BuSl#login/"
A couple of things I have already tried
^.{5}\Kstate
But seems not working. Any ideas. I need this to retrieve for jMeter testing.
No need of lookbehind, nor any lookarounds at all. Use a single capturing group and a negated character class:
state%3([^#]+)
AND set the template value to $1$.
See the regex demo. Details:
state%3 - matches a literal text
([^#]+) - Capturing group #1 (that is why template should be $1$): one or more chars other than #.
I have a string containing the following variable "nonce=1ff7de7518b9a52080489ecd7629796d&" how to get the value between the equal and the "&" in regular expression, I have tried nonce=(.*?).+?(?=&) the ending part excluded "&" but I could not exclude "nonce="
Note: trying to match the value between "=" and "&" will not work as there are many "=" and "&" characters which will result in more than 1 match, the unique string is "nonce"
here is an example https://regexr.com/48vmd
You can use nonce=([^&]+) to match and capture your intended string from group1
Here nonce= will match literally and then ([^&]+) will match all text before & and capture in group1.
Demo
In case your regex flavor supports \K match reset operator, you can use this regex nonce=\K[^&]+ to have your intended text as full match without requiring any group text capture.
Demo without any grouped capture
If you're using Java, you can use this regex which uses look behind and Java supports look behind.
(?<=nonce=)[^&]+
Demo using look behind
If you're looking for the regular expression it would be as simple as nonce=(\w+)&
Demo (assumes RegExp Tester mode of the View Results Tree listener)
Even easier way would be going for Boundary Extractor which basically extracts everything between the given "left" and "right" boundaries:
Is there a way to write regular expressions to stop right before a particular word or characters?
For example, I have a text like:
Advisor:HarrisTeamTeamRole
So I want to write a regular expression that makes the advisor name dynamic, but only capture Harris. How do I write a regular expression to stop right before Team?
You could use a lookbehind and lookahead like this:
(?<=Advisor:).*?(?=Team)
Debuggex Demo
This will only capture from "Advisor:" up to the first "Team", and the regex will not capture anything else after (including "Team") in a capture group or otherwise. This will require a type of regex that can do lookbehinds... if you are not using that, you'll have to use grouping... which could be as simple as:
Advisor:(.*?)Team
and then just get the capture group #1
Try this one
This regular expression would be:
:([A-Z][a-z]*)
This one captures only the first word after the colon as long as it's in CamelCase, meaning it doesn't have to be the word Team it could be Advisor:HarrisNetworkSomething as well.
You can try in Lazy way and get the matched group from index 1
^Advisor:(.*?)Team
Here is online demo
I have the sentence as below:
First learning of regular expression.
And I want to extract only First learning and expression by means of regular expressions.
Where would I start/
Regular expressions are for pattern matching, which means we'd need to know a pattern that is to be matched.
If you literally just want those strings, you'd just use First learning and expression as your patterns.
As #orique says, this is kind of pointless; you don't need RegEx for that. If you want something more complicated, you'd need to explain what you're trying to match.
Regex is not usually used to match literal text like what you're doing, but instead is used to match patterns of text. If you insist on using regex, you'll have to match the trivial expression
(First learning|expression)
As already pointed out, it is unusual to match a literal string like you are asking, but more common to match patterns such as several word characters followed by a space character etc...
Here is a pattern to match several word characters (which are a-z, A-Z, 0-9 and _) followed by a space, followed by several more word characters etc... It ends up capturing three groups. The first group will match the first two words, the second part the next to words, and the last part, the fifth word and the preceding space.
$words = "First learning of regular expression.";
preg_match(/(\w+\s\w+)\s(\w+\s\w+)(\s\w+)/, $words, $matches);
$result = matches[1]+matches[3];
I hope this matches your requirement.