I've got a single route function that I want to match two different paths to. The idea is to create profile pages for team members of a website. They'll be at www.domain.com/name. The issue is, some people have nicknames. So, I need www.domain.com/nickname to go to the same place as www.domain.com/name.
Here's what I've got so far:
website.get('/name|nickname', websiteRoutes.about);
The problem is things like /asdasdfdnickname and /namezzzzzzzz will match as well.
How do I match either the name or nickname only without any extra characters. I believe this is called an exclusive or?
So here are some working solutions
Passing in ['/name', '/nickname'] into the routing function.
And from John's answer below: /^\/?(name|nickname)\/?$/i
Try /^\/?(name|nickname)\/?$/i which will match exactly name/nickname only.
This regex means it can optionally start with a forward slash, it will match "name" or "nickname" case insensitively, then it will optionally allow another forward slash at the end.
Related
I try to figure out the correct regex to replace the last segment of an url with a modified version of that very last segment. (I know that there are similar threads out there, but none seemed to help...)
Example:
https://www.test.com/one/two/three/mypost/
--->
one/two/three?id=mypost
https://www.test.com/one/mypost/
--->
one?id=mypost
Now I am stuck here:
https://regex101.com/r/9GqYaU/1
I can get the last segment in capturing group 2 but how would I replace it?
I think I will have to something like this:
const url = 'https://www.test.com/one/two/three/mypost/'
const regex = /(http[s]?:\/\/)([^\/]+\/)*(?=\/$|$)/
const path = url.replace(regex, `${myUrlWithoutTheLastSegmentAnd WithoutHTTPS}?id=$2`)
return path
But I have no idea how to get the url without the last segment. I have currently only access to the whole string or group 1 (which is useless in this case) and then group 2, but not the string without group 2.
I would be very glad for any help here. Sometimes I just lack the knowledge of what is possible with regex and how to achieve it.
Thank you in advance.
Cheers
You could use the URL class to extract the pathname and substring to remove the first '/'.
Then, you could put the last part of the pathname in a group and use it as a reference $1 for the replacement.
const url = new URL('https://www.test.com/one/two/three/mypost/').pathname.substring(1)
console.log(url.replace(/\/([^/]*)\/$/, '?id=$1'))
I came across your question yesterday and agree with going down the route of parsing the URL. Once you get there you could even use JavaScript array methods which I prefer to string methods like:
pathname.split("/").filter(p => p.length).pop()
This would separate each folder, ignore any with no length (i.e. handle a trailing slash) and return the last one (mypost).
Anyway, I am also learning regex so sometimes when I find a question like this I just try to find the answer anyway as the best way of learning is doing. It took 24 hours 😂 I came up with this:
/(https?:\/\/).+?([a-z-]*)\/?$/gm
(https?:\/\/) you know what this does. Small correction, you don't need the square brackets. Question mark matches 0 or 1 of the preceding character. As we're only matching s this just works. If you wanted to match s or z you would use [sz]?. I think.
.+? this is the cool one I think I will use in future now I found it. The question mark here has a different meaning - it makes .+ (which means one or more of any character) non-greedy. That means it stops applying once it reaches the next rule. Which is...
([a-z-]*) any number of letters or a hyphen. You should maybe change this to include numbers and upper case.
\/? Optional slash
$ all this must apply at the end of the string.
Here is a demo
https://regex101.com/r/mQNkIS/1
I've racked my brain over this JS regex and have so far only managed to get parts of it to work or the whole thing to work in certain circumstances.
I have a string like this:
Some string<br>http://anysubdomain.particulardomain.com<br>Rest of string
The goal is to move the domain part to the end of the string, if it's there. The http part is also optional and can also be https. The TLD is always particulardomain.com, the subdomain can be anything.
I've managed to get everything into capture groups when the domain with protocol is present with this regex:
(.*)(https?\:\/\/[a-z\d\-]*\.particulardomain\.com)(.*)
But any attempt at making the domain part and the protocol part within it optional has resulted in no or the wrong matches.
The end result I'm looking for is to have the three parts of the string – beginning, domain, end – in separate capture groups so I can move capture group 2 (the domain part) to the end, or, if there's no domain present, the whole string in the first capture group.
To clarify, here are some examples with the expected output/capture groups:
INPUT:
Some string<br>http://anysubdomain.particulardomain.com<br>Rest of string
OR (no protocol):
Some string<br>anysubdomain.particulardomain.com<br>Rest of string
OUTPUT:
$1: Some string<br>
$2: http://anysubdomain.particulardomain.com
$3: <br>Rest of string
INPUT:
Some string<br>Rest of string
OUTPUT:
$1: Some string<br>Rest of string
$2: empty
$3: empty
One mistake in your regex is that it contains only particular whereas
the source text contains particulardomain, but this is a detail.
Now let's move to the protocol part. You put only one ? (after s),
which means that only s is optional, but both http and :
are still required.
To make the whole protocol optional, you must:
enclose it with a group (either capturing or not),
make this group optional (put ? after it).
And now maybe the most important thing: Your regex starts with (.*).
Note that it is greedy version, which:
initially tries to capture the whole rest of source string,
then moves back one char by one, to allow matching by the
following part of regex.
Change it to reluctant version (.*?) and then optional
group (https?:)? will match as expected.
Another detail: \ before : is not needed. It does not do
any harm either, but due to the principle "Keep It Simple...",
I recommend to delete it (as I did above).
One more detail: After [a-z\d\-] (subdomain part) you should put
+, not *, as this part may not be empty.
So the whole regex can be:
(.*?)((https?:)?\/\/[a-z\d\-]+\.particulardomain\.com)(.*)
And the last remark: I am in doubt, whether you really need three
capturing groups. Maybe it would be enough to leave only the content
of the middle capturing group, i.e.:
(https?:)?\/\/[a-z\d\-]+\.particulardomain\.com
Found a solution. Since, as stated, the goal is to move the domain to the end of the string, if it's present, I'm just matching the domain and anything after it. If there's no domain, nothing matches and hence nothing gets replaced. The problem was the two .* both at the beginning and the end of the regex. Only the one at the end is needed.
REGEX:
([a-z\d\-:\/]+\.particulardomain\.com)(.*)
Works for the following strings:
Domain present:
Start of string 1234<br>https://subdomain.particulardomain.com<br>End of string 999
Domain without protocol:
Start of string 1234<br>subdomain.particulardomain.com<br>End of string 999
No domain:
Start of string 1234<br>End of string 999
Thanks everyone for helping me rethink the problem!
I see good answer here, as you explained you need three group and set the domain to the back of the string(to be clear the entire url or only the domain e.g particulardomain.com)
You can do this:
//Don't know if the <br> tag matter for you problem, suppose it not
//this is you input
let str = "Start of string 1234<br>https://subdomain.particulardomain.com<br>End of string 99";
let group = str.split(<br>);
let indexOfDomain;
/*moere code like a for loop or work with a in-build funcion of the array with the regExp you made /[a-z\d\-:\/]+\.particulardomain\.com/ you can validated the domain separately.
}
TO HAVE IN MIND:
With your solution will not work at 100%, why?
your regExp:
([a-z\d\-:\/]+\.particulardomain\.com)(.*)
will mach a http, https, *(any other thing that is not a protocol) and will not work for this input you can test if you like and do a comment
Start of string 1234<br>End of string 999
The regExp that #Valdi_Bo answer:
(.*?)((https?:)?\/\/[a-z\d\-]+\.particulardomain\.com)(.*)
will fit to the what you described in the question
This regExp don't fit all yours input maybe he did not test it for all your input as you did not explained in your question like you did in your own answer
In conclusion at the end you need to extract the domain (wich don't know if is the entire url as you mix up the idea). If you are not going to use the do a split and then validated the regExp it will be more easy
Consider this URL:
domains.google.com/registrar#t=b
note:
#t=b
In this example, the variable "t" stores the current tab on the page where "b" is for billing.
How can I achieve query like parameters in backbone as shown above?
I understand Backbone has routes that support parameters in urls, but this is limited to when the data is in a hierarchy, for example: item/:id
But what about application settings that would not work well in a directory like structure?
The only solution I can think of is a custom parser and break up the key/values myself.
Any ideas?
Expanding on #try-catch-finally's comment, I'm going to show you how to create your own route with a simple RegEx pattern that will match your conditions.
Here's the regex we'll use:
^\w+? # match one word (or at least a character) at the
# beginning of the route
[=] # until we parse a single 'equals' sign
( # then start saving the match inside the parenthesis
[a-zA-Z0-9]* # which is any combination of letters and numbers
) # and stop saving
Putting it all together the regex looks like: /^\w+?[=]([a-zA-Z0-9]*)/.
Now we set up our router,
var MyRouter = Backbone.Router.extend({
initialize: function(options) {
// Matches t=b, passing "b" to this.open
this.route(/^\w+?(?<=[=])(.*)/, "testFn");
},
routes: {
// Optional additional routes
},
testFn: function (id) {
console.log('id: ' + id );
}
});
var router = new MyRouter();
Backbone.history.start();
The TL;DR is that in the MyRouter.initialize we added a route that takes the regex above and invokes the MyRouter.testFn function. Any call to http://yourdomain.com#word1=word2 will call the MyRouter.testFn with the word after the parenthesis as a parameter. Of course, your word place setting could be a single character like in your question t=b.
Expanding your parameters
Say you want to pull multiple parameters, not just the one. The key to understanding how Backbone pulls your parameters is the capturing group (). A capturing group allows your to "save" the match defined within the parenthesis into variables local to the regex expression. Backbone uses these saved matches as the parameters it applies to the the route callback.
So if you want to capture two parameters in your route you'd use this regex:
^\w+?[=]([a-zA-Z0-9]*)[,]\w+?[=]([a-zA-Z0-9]*)
which simply says to expect a comma delimiter between the two parameter placeholders. It would match,
t=b,some=thing
More general route patterns
You can repeat the [,]\w+?[=]([a-zA-Z0-9]*) pattern as many times as you need. If you want to generalize the pattern, you cold use the non-capturing token (?: ... ) and do something like,
^\w+?[=]([a-zA-Z0-9]*)(?:[,]\w+?[=]([a-zA-Z0-9]*))?(?:[,]\w+?[=]([a-zA-Z0-9]*))?
The regex above will look for at least one match and will optionally take two more matches. By placing a ? token at the end of the (?: ... ) group, we say the pattern in the parenthesis may be found zero or one times (i.e. it may or may not be there). This allows you to set a route when you know you can expect up to 3 parameters, but sometimes you may want only one or two.
Is there a truly general route?
You may be asking yourself, why not simply use one greedy (?: ... ) group that will allow an unlimited number of matches. Something like,
^\w+?[=]([a-zA-Z0-9]*)(?:[,]\w+?[=]([a-zA-Z0-9]*))*
With this regex pattern you must supply one parameter, but you can take an unlimited number of subsequent matches. Unfortunately, while the regex will work fine, you won't get the desired result. (See, for example, this Question.)
That's a limitation of JavaScript. With repeating capturing-groups (i.e. the ([a-zA-Z0-9]*) capturing-group will repeat with every repetition of the (?: ... ) non-capturing-group) JavaScript only saves the last match. So if you pass a route like t=b,z=d,g=f,w=1ee, you'll only save 1ee. So, unfortunately you have to have an idea of what the maximum number of parameters your route should take, and manually code them into your regex pattern like we did above.
While writing an API service for my site, I realized that String.split() won't do it much longer, and decided to try my luck with regular expressions. I have almost done it but I can't find the last bit. Here is what I want to do:
The URL represents a function call:
/api/SECTION/FUNCTION/[PARAMS]
This last part, including the slash, is optional. Some functions display a JSON reply without having to receive any arguments. Example: /api/sounds/getAllSoundpacks prints a list of available sound packs. Though, /api/sounds/getPack/8Bit prints the detailed information.
Here is the expression I have tried:
req.url.match(/\/(.*)\/(.*)\/?(.*)/);
What am I missing to make the last part optional - or capture it in whole?
This will capture everything after FUNCTION/ in your URL, independent of the appearance of any further / after FUNCTION/:
FUNCTION\/(.+)$
The RegExp will not match if there is no part after FUNCTION.
This regex should work by making last slash and part after optional:
/^\/[^/]*\/[^/]*(?:\/.*)?$/
This matches all of these strings:
/api/SECTION/FUNCTION/abc
/api/SECTION
/api/SECTION/
/api/SECTION/FUNCTION
Your pattern /(.*)/(.*)/?(.*) was almost correct, it's just a bit too short - it allows 2 or 3 slashes, but you want to accept anything with 3 or 4 slashes. And if you want to capture the last (optional) slash AND any text behind it as a whole, you simply need to create a group around that section and make it optional:
/.*/.*/.*(?:/.+)?
should do the trick.
Demo. (The pattern looks different because multiline mode is enabled, but it still works. It's also a little "better" because it won't match garbage like "///".)
I 'borrowed' a regex from this website : http://daringfireball.net/2010/07/improved_regex_for_matching_urls that is almost complete but i want to match exemple.com
I know that stackoverflow is not doyourhomework.com but I passed a long time thinking without results. Here is a fiddle to test : http://jsfiddle.net/BGnMm/25/ and you can see at the end that exemple.com is not a link.
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}\/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
var allurl="http:foo.com/blah_blah http://foo.com/blah_blah/ (Something like http://foo.com/blah_blah) http://foo.com/blah_blah_(wikipedia) http://foo.com/more_(than)_one_(parens) (Something like http://foo.com/blah_blah_(wikipedia)) http://foo.com/blah_(wikipedia)#cite-1 http://foo.com/blah_(wikipedia)_blah#cite-1 http://foo.com/unicode_(✪)_in_parens http://foo.com/(something)?after=parens http://foo.com/blah_blah. http://foo.com/blah_blah/. <http://foo.com/blah_blah> <http://foo.com/blah_blah/> http://foo.com/blah_blah, http://www.extinguishedscholar.com/wpglob/?p=364. http://✪df.ws/1234 rdar://1234 rdar:/1234 x-yojimbo-item://6303E4C1-6A6E-45A6-AB9D-3A908F59AE0E message://%3c330e7f840905021726r6a4ba78dkf1fd71420c1bf6ff#mail.gmail.com%3e http://➡.ws/䨹 www.c.ws/䨹 <tag>http://example.com</tag> Just a www.example.com link. http://example.com/something?with,commas,in,url, but not at end What about <mailto:gruber#daringfireball.net?subject=TEST> (including brokets). mailto:name#example.com bit.ly/foo “is.gd/foo/” WWW.EXAMPLE.COM http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55))/Web_ENG/View_DetailPhoto.aspx?PicId=752 http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55)) http://lcweb2.loc.gov/cgi-bin/query/h?pp/horyd:#field(NUMBER+#band(thc+5a46634)) 6:00p filename.txt http://example.com/quotes-are-“part” ✪df.ws/1234 example.com example.com/";
document.write(allurl.replace(reg,"<a href='$1' >$1</a><br />"));
Add an alternation operator (|) after the {2,4}\/, i.e.
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}\/|)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
There's something you should understand about this. The first non-captured group, (?: … ), looks for "indicators" of URLs. One indicator, for example, is the www (followed by up to 3 digits of numbers). You however are asking for a way to identify URLs without any indicator at all. So, what we've done above is we've added a clause, "or an empty match," as a "valid" indicator. The consequence of this is that your regular expression is less selective now: all sorts of strings, not only example.com but also filename.txt, 3.141593, and omg...really are identified as URLs! Your only other (readily available) option is to be more selective about suffixes, e.g. require specific suffixes (com|org|net), but then this takes away from the generality of the original regex, which doesn't specify any suffixes at all.
In other words, you are probably faced with a limitation of logic, not a limitation of regex-writing skills or the regex language itself.
Please check if
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|(?:www\d{0,3}[.])|[a-z0-9.\-]+[.][a-z]{2,4}\/{0,1})(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))*(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
suits your needs. www(anyNumber) has just been put to appear one or zero times. Sorry for the first answer, did not notice the texts.