I have lines like this:
example.com/p/stuff/...
example.com/page/thing/...
example.com/page/stuff/...
example.com/page/other-stuff/...
etc
where the dots represent continuing URL paths. I want to select URLs that contain /page/ and are NOT followed by thing/. So from the above list we would select:
example.com/page/stuff/...
example.com/page/other-stuff/...
.*?\/page\/[^(thing)].*
this is the regex for matching a string which has /page/ not followed by thing
adding the lazy evalation is suggested because you advance a char at the time, better performance!
You need to use negative lookahead:
example\.com\/page\/(?!thing\/).*
Demo
Use the following regex pattern:
.*?\/page\/(?!thing\/).*
https://regex101.com/r/19wh1w/2
(?!thing\/) - negative lookahead assertion ensures that page/ section is not followed by thing/
Related
RegEx to match one of two strings in the third segment, ie in pseudo code:
/content/au/(boomer or millenial)/...
Example matches
/content/au/boomer
/content/au/boomer/male/31
/content/au/millenial/female/29/M
/content/au/millenial/male/18/UM
Example non-matches
/content/au
/content/nz/millenial/male/18/UM
/content/au/genz/male
I've tried this, but to no avail:
^/content/au/(?![^/]*/(?:millenial|boomer))([^/]*)
Don't use a look ahead; just use the plain alternation millenial|boomer then a word-boundary:
^/content/au/(?:millenial|boomer)\b(?:/.*)?
See live demo.
You should probably spell millennial correctly too (two "n"s, not one).
What's with the negative lookahead? This is a simple, if not trivial, positive match.
^/content/au/(?:millenial|boomer)(?:/|$)
The final group says the match needs to be followed by a slash or nothing, so as to exclude paths which begin with one of the alternatives, but contain additional text.
You can use the following regex DEMO
content/au/(?:boomer|millenial)
I need to create a route that responses to any string starting with '#' character. Routes like following examples :
www.mywebsite.com/#john
www.mywebsite.com/#jack
www.mywebsite.com/#something
So I wrote:
Route::get('{something}','SomeController#someMethod')->where('something','/#^/');
But when I test it, I face 404 not found found page.
what is the correct regular expression for this?
Route::get('/{tag}', 'SomeController#someMethod')->where('tag', '^#.*');
This will also work:
Route::get('#{something}', 'SomeController#someMethod');
You can write this
Route::pattern('tag', '#[a-zA-Z]');
Route::get('{tag}', 'SomeController#someMethod');
This way you seperate the logic of the regex and the route and it will work as you want
Note the #^ pattern means # should be followed with the beginning of string, which is not possible, and the pattern never matches any string. The '^#' pattern asserts the position at the start of the string, and only there does it try to match #.
Also, the usual / regex delimiters should be removed from this pattern as they are treated as part of the pattern here.
So, in your case you may just swap the anchor and the # char:
Route::get('{something}','SomeController#someMethod')->where('something','^#');
Consider the following input string (part of css file):
url('data:image/png;base64,iVBORw0KGgoAAAAN...');
url(example.png);
The objective is to take the url part using regex and do something with it. So the first part is easy:
url\(['"]?(.+?)['"]?\)
Basically, it takes contents from inside url(...) with optional quotes symbols. Using this regexp I get the following matches:
data:image/png;base64,iVBORw0KGgoAAAAN...
example.png
So far so good. Now I want to exclude the urls which include 'data:image' in their text. I think negative lookahead is the proper tool for that but using it like this:
url\(['"]?(?!data:image)(.+?)['"]?\)
gives me the following result for the first url:
'data:image/png;base64,iVBORw0KGgoAAAAN...
Not only it doesn't exclude this match, but the matched string itself now includes quote character at the beginning. If I use + instead of first ? like this:
url\(['"]+(?!data:image)(.+?)['"]?\)
it works as expected, url is not matched. But this doesn't allow the optional quote in url (since + is 1 or more). How should I change the regex to exclude given url?
You can use negative lookahead like this:
url\((['"]?)((?:(?!data:image).)+?)\1?\)
RegEx Demo
Ok, I know that it is a question often asked, but I did not manage to get what I wanted.
I am looking for a regular expression in order to find a pattern that does not contain a particular substring.
I want to find an url that does not contains the b parameter.
http://www.website.com/a=789&c=146 > MATCH
http://www.website.com/a=789&b=412&c=146 > NOT MATCH
Currently, I have the following Regex:
\bhttp:\/\/www\.website\.com\/((?!b=[0-9]+).)*\b
But I am wrong with the \b, the regex match the beginning of th string and stop when it find b=, instead of not matching.
See: http://regex101.com/r/fN3zU5/3
Can someone help me please?
Just use a lookahead to check anything following the URL must be a space or line end.
\bhttp:\/\/www\.website\.com\/(?:(?!b=[0-9]+).)*?\b(?= |$)
DEMO
use this:
^http:\/\/www\.website\.com\/((?!b=[0-9]+)).*$
\b only matches word endings.
^ matches start and end of string
and you dont even need to do it that complicated, If you dont want the url with the b parameter use this:
^http:\/\/www\.website\.com\/(?!b).*$
demo here : http://regex101.com/r/fN3zU5/5
import re
pattern=re.compile(r"(?!.*?b=.*).*")
print pattern.match(x)
This will look ahead if there is a "b=" present.A negative lookahead means it will not match that string.
You had a look at this possibility:
http://regex101.com/r/fN3zU5/6
^http:\/\/www\.website\.com\/[ac\=\d&]*$
only allow &,=,a,c and digits
complete url in group and there should not be a "b=" parameter
if you have more options and you dont want to list them all:
you dont allow a 'b' to be part of your parameters
^http:\/\/www\.website\.com\/[^b]*$
http://regex101.com/r/fN3zU5/7
^http:\/\/www\.website\.com\/(?!.*?b=.*?).*$ works too here "b=" is permitted at any position of the parameter string so you could even have the "b" string as a value of a parameter.
See
http://regex101.com/r/fN3zU5/8
This is what you want. ^http:\/\/www\.website\.com\/(([^b]=[0-9]+).)*$
Its a simple pattern not flexible but it works :
http:\/\/www\.website\.com\/+a=+\w+&+c=+\w+
I'm struggling to come up with the correct regex for the following scenario.
Let's say you have to match a word either starts with http- or nothing
eg : http-test-data, test-data should be a match but xyz-test-data shouldn't be a match
the regex i came up so far is
(?:http-)?(test-data)
but it matches xyz-test-data as well.
You could simply use the following:
(?:http-|^)(test-data)
This tests for either a positive look-behind of http- or for the beginning of the string before test-data.
For example, for the sample data as follows:
http-test-data
xyz-test-data
http-test-data
xyz-test-data
test-data
yes-yes-test-data
-test-data
It yeilds:
http-test-data
http-test-data
test-data
Try this representation
^(http-|)(test-data)
Yes because there is a ? on the (?:http-). Then the regex will also match any string that contains test-data.