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:
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" ]
my regular expression :(?<=defining\s)[^;]*
should try to find var in the following cases:
defining var;
some text defining var;
I tested the regular expression using some online tools. Unfortunately it does not work with Qt for some reason. Is there something wrong with the regular expression or should I look for the error somewhere else in the code? It is strange because regular expressions without lookbehind work.
Extra:
a bit off topic but because I'm already writing the question: How do I have to change my regular expression so that it can find var in the following case with more then one whitespace:
some text defining var;
Right now it finds all but one whitespace and var.
You can match the context to the left of the value you want to obtain and capture the latter into a capturing group:
QRegularExpression regex("defining\\s+([^;]+)");
QRegularExpressionMatch match = regex.match(str);
QString textYouWant = match.captured(1);
Here, defining\\s+([^;]+) matches defining, then 1+ whitespace chars, and then captures 1+ chars other than ; into Group 1 (that you can access using .captured(1)).
See this regex demo.
Note that QRegularExpression is PCRE-powered, so you may use the PCRE \K operator to get the value you need in the match itself:
QRegularExpression regex("defining\\s+\\K[^;]+");
^^^
The \K operator discards the text matched so far, so you will only get the text matched with [^;]+ after the match is returned.
See another regex demo.
I'd like to know how can I ignore characters that follows a particular pattern in a Regex.
I tried with positive lookaheads but they do not work as they preserves those character for other matches, while I want them to be just... discarded.
For example, a part of my regex is: (?<DoubleQ>\"\".*?\"\")|(?<SingleQ>\".*?\")
in order to match some "key-parts" of this string:
This is a ""sample text"" just for "testing purposes": not to be used anywhere else.
I want to capture the entire ""sample text"", but then I want to "extract" only sample text and the same with testing purposes. That is, I want the group to match to be ""sample text"", but then I want the full match to be sample text. I partially achieved that with the use of the \K option:
(?<DoubleQ>\"\"\K.*?\"\")|(?<SingleQ>\"\K.*?\")
Which ignores the first "" (or ") from the full match but takes it into account when matching the group. How can I ignore the following "" (")?
Note: positive lookahead does not work: it does not ignore characters from the following matches, it just does not include them in the current match.
Thanks a lot.
I hope I got your questions right. So you want to match the whole string including the quotes, but you want to replace/extract it only the expression without the quotes, right?
You typically can use the regex replace functionality to extract just a part of the match.
This is the regex expression:
""?(.*?)""?
And this the replace expression:
$1
Is there a way to write regular expressions to stop right before a particular word or characters?
For example, I have a text like:
Advisor:HarrisTeamTeamRole
So I want to write a regular expression that makes the advisor name dynamic, but only capture Harris. How do I write a regular expression to stop right before Team?
You could use a lookbehind and lookahead like this:
(?<=Advisor:).*?(?=Team)
Debuggex Demo
This will only capture from "Advisor:" up to the first "Team", and the regex will not capture anything else after (including "Team") in a capture group or otherwise. This will require a type of regex that can do lookbehinds... if you are not using that, you'll have to use grouping... which could be as simple as:
Advisor:(.*?)Team
and then just get the capture group #1
Try this one
This regular expression would be:
:([A-Z][a-z]*)
This one captures only the first word after the colon as long as it's in CamelCase, meaning it doesn't have to be the word Team it could be Advisor:HarrisNetworkSomething as well.
You can try in Lazy way and get the matched group from index 1
^Advisor:(.*?)Team
Here is online demo
I've got (To) [a-z]+ as regular expression and I've got sentence: To kot dziki pies.
And If I compile it I will retrieve To kot.
So what can I do to retrieve only word after (only kot) "To" instead of "To kot"?
^To (\w+)$ 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.