I'm trying to create a generic regex pattern for a crawler, to avoid so called "crawler traps" (links that just add url parameters and refer to the exact same page, which results in tons of useless data). Alot of times, those links just add the same part to a URL over and over again. Here is an example out of a log file:
http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/...
I can use regular expressions to narrow the scope of the crawler and i would love to have a pattern, that tells the crawler to ignore everything that has repeating parts. Is that possible with a regex?
Thanks in advance for some tips!
JUST TO CLARIFY:
the crawlertraps are not designed to prevent crawling, they are a result of poor web design. All the pages we are crawling explicitly allowed us to do so!
If you are already looping through a list of URLs, you could add matching as a condition to skip the current iteration:
array = ["/abcd/abcd/abcd/abcd/", "http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/", "http://examplepage/apple/cake/banana/"]
import re
pattern1 = re.compile(r'.*?([^\/\&?]{4,})(?:[\/\&\?])(.*?\1){3,}.*')
for url in array:
if re.match(pattern1, url):
print "It matches; skipping this URL"
continue
print url
Example regex:
.*?([^\/\&?]{4,})(?:[\/\&\?])(.*?\1){3,}.*
([^\/\&?]{4,}) matches and captures sequences of anything, but not containing [/&?], repeated 4 or more times.
(?:[\/\&\?]) looks for one /,& or ?
(.*?(?:[\/\&\?])\1){3,} match anything until [/&?], followed by what we captured, doing all of this 3 or more times.
demo
You can use a backreference in Python/PERL regexes (and possibly others) to catch a pattern which is repeated:
>>> re.search(r"(/.+)\1", "http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/").group(1)
'/cssms/chrome'
\1 references the first match, so (/.+)\1 means the same sequence repeated twice in a row. The leading / is just to avoid the regex matching the first single repeating letter (which is the t in http) and catch repetitions in the path.
Related
I'm using the following regex to find URLs in a text file:
/http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+/
It outputs the following:
http://rda.ucar.edu/datasets/ds117.0/.
http://rda.ucar.edu/datasets/ds111.1/.
http://www.discover-earth.org/index.html).
http://community.eosdis.nasa.gov/measures/).
Ideally they would print out this:
http://rda.ucar.edu/datasets/ds117.0/
http://rda.ucar.edu/datasets/ds111.1/
http://www.discover-earth.org/index.html
http://community.eosdis.nasa.gov/measures/
Any ideas on how I should tweak my regex?
Thank you in advance!
UPDATE - Example of the text would be:
this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/).
This will trim your output containing trail characters, ) .
import re
regx= re.compile(r'(?m)[\.\)]+$')
print(regx.sub('', your_output))
And this regex seems workable to extract URL from your original sample text.
https?:[\S]*\/(?:\w+(?:\.\w+)?)?
Demo,,, ( edited from https?:[\S]*\/)
Python script may be something like this
ss=""" this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/). """
regx= re.compile(r'https?:[\S]*\/(?:\w+(?:\.\w+)?)?')
for m in regx.findall(ss):
print(m)
So for the urls you have here:
https://regex101.com/r/uSlkcQ/4
Pattern explanation:
Protocols (e.g. https://)
^[A-Za-z]{3,9}:(?://)
Look for recurring .[-;:&=+\$,\w]+-class (www.sub.domain.com)
(?:[\-;:&=\+\$,\w]+\.?)+`
Look for recurring /[\-;:&=\+\$,\w\.]+ (/some.path/to/somewhere)
(?:\/[\-;:&=\+\$,\w\.]+)+
Now, for your special case: ensure that the last character is not a dot or a parenthesis, using negative lookahead
(?!\.|\)).
The full pattern is then
^[A-Za-z]{3,9}:(?://)(?:[\-;:&=\+\$,\w]+\.?)+(?:\/[\-;:&=\+\$,\w\.]+)+(?!\.|\)).
There are a few things to improve or change in your existing regex to allow this to work:
http[s]? can be changed to https?. They're identical. No use putting s in its own character class
[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),] You can shorten this entire thing and combine character classes instead of using | between them. This not only improves performance, but also allows you to combine certain ranges into existing character class tokens. Simplifying this, we get [a-zA-Z0-9$-_#.&+!*\(\),]
We can go one step further: a-zA-Z0-9_ is the same as \w. So we can replace those in the character class to get [\w$-#.&+!*\(\),]
In the original regex we have $-_. This creates a range so it actually inclues everything between $ and _ on the ASCII table. This will cause unwanted characters to be matched: $%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_. There are a few options to fix this:
[-\w$#.&+!*\(\),] Place - at the start of the character class
[\w$#.&+!*\(\),-] Place - at the end of the character class
[\w$\-#.&+!*\(\),] Escape - such that you have \- instead
You don't need to escape ( and ) in the character class: [\w$#.&+!*(),-]
[0-9a-fA-F][0-9a-fA-F] You don't need to specify [0-9a-fA-F] twice. Just use a quantifier like so: [0-9a-fA-F]{2}
(?:%[0-9a-fA-F][0-9a-fA-F]) The non-capture group isn't actually needed here, so we can drop it (it adds another step that the regex engine needs to perform, which is unnecessary)
So the result of just simplifying your existing regex is the following:
https?://(?:[$\w#.&+!*(),-]|%[0-9a-fA-F]{2})+
Now you'll notice it doesn't match / so we need to add that to the character class. Your regex was matching this originally because it has an improper range $-_.
https?://(?:[$\w#.&+!*(),/-]|%[0-9a-fA-F]{2})+
Unfortunately, even with this change, it'll still match ). at the end. That's because your regex isn't told to stop matching after /. Even implementing this will now cause it to not match file names like index.html. So a better solution is needed. If you give me a couple of days, I'm working on a fully functional RFC-compliant regex that matches URLs. I figured, in the meantime, I would at least explain why your regex isn't working as you'd expect it to.
Thanks all for the responses. A coworker ended up helping me with it. Here is the solution:
des_links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', des)
for i in des_links:
tmps = "/".join(i.split('/')[0:-1])
print(tmps)
I have two URLs with the patterns:
1.http://localhost:9001/f/
2.http://localhost:9001/flight/
I have a site filter which redirects to the respective sites if the regex matches. I tried the following regex patterns for the 2 URLs above:
http?://localhost[^/]/f[^flight]/.*
http?://localhost[^/]/flight/.*
Both URLS are getting redirected to the first site, as both URLs are matched by the first regex.
I have tried http?://localhost[^/]/[f]/.* also for the 1st url. I am Unable to get what am i missing . I feel that this regex should not accept any thing other than "f", but it is allowing "flight" as well.
Please help me by pointing the mistake i have done.
Keep things simple:
.*/f(/[^/]*)?$
vs
.*/flight(/[^/]*)?$
Adding ? before $ makes the trailing slash with optional path term optional.
The first one will be caught with following regex;
/^http:[\/]{2}localhost:9001\/f[^light]$/
The other one will be disallowed and can be found with following regex
/^http:[\/]{2}localhost:9001\/flight\/$/
You regex has several issues: 1) p? means optional p (htt:// will match), 2) [^/] will only match : in your URLs since it will only capture 1 character (and you have a port number), 3) [^light] is a negated character class that means any character that is not l, i, g, h, or t.
So, if you want to only capture localhost URLs, you'd better use this regex for the 1st site:
http://localhost[^/]*/f/.*
And this one for the second
http://localhost[^/]*/flight/.*
Please also bear in mind that depending on where you use the regexps, your actual input may or may not include either the protocol.
These should work for you:
http[s]{0,1}:\/\/localhost:[0-9]{4}\/f\/
http[s]{0,1}:\/\/localhost:[0-9]{4}\/flight\/
You can see it working here
How can I extract only top-level and second-level domain from a URL using regex? I want to skip all lower level domains. Any ideas?
Here's my idea,
Match anything that isn't a dot, three times, from the end of the line using the $ anchor.
The last match from the end of the string should be optional to allow for .com.au or .co.nz type of domains.
Both the last and second last matches will only match 2-3 characters, so that it doesn't confuse it with a second-level domain name.
Regex:
[^.]*\.[^.]{2,3}(?:\.[^.]{2,3})?$
Demonstration:
Regex101 Example
Updated 2019
This is an old question, and the challenge here is a lot more complicated as we start adding new vanity TLDs and more ccTLD second level domains (e.g. .co.uk, .org.uk). So much so, that a regular expression is almost guaranteed to return false positives or negatives.
The only way to reliably get the primary host is to call out to a service that knows about them, like the Public Suffix List.
There are several open-source libraries out there that you can use, like psl, or you can write your own.
Usage for psl is quite intuitive. From their docs:
var psl = require('psl');
// Parse domain without subdomain
var parsed = psl.parse('google.com');
console.log(parsed.tld); // 'com'
console.log(parsed.sld); // 'google'
console.log(parsed.domain); // 'google.com'
console.log(parsed.subdomain); // null
// Parse domain with subdomain
var parsed = psl.parse('www.google.com');
console.log(parsed.tld); // 'com'
console.log(parsed.sld); // 'google'
console.log(parsed.domain); // 'google.com'
console.log(parsed.subdomain); // 'www'
// Parse domain with nested subdomains
var parsed = psl.parse('a.b.c.d.foo.com');
console.log(parsed.tld); // 'com'
console.log(parsed.sld); // 'foo'
console.log(parsed.domain); // 'foo.com'
console.log(parsed.subdomain); // 'a.b.c.d'
Old answer
You could use this:
(\w+\.\w+)$
Without more details (a sample file, the language you're using), it's hard to discern exactly whether this will work.
Example: http://regex101.com/r/wD8eP2
Also, you can likely do that with some expression similar to,
^(?:https?:\/\/)(?:w{3}\.)?.*?([^.\r\n\/]+\.)([^.\r\n\/]+\.[^.\r\n\/]{2,6}(?:\.[^.\r\n\/]{2,6})?).*$
and add as much as capturing groups that you want to capture the components of a URL.
Demo
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
For anyone using JavaScript and wanting a simple way to extract the top and second level domains, I ended up doing this:
'example.aus.com'.match(/\.\w{2,3}\b/g).join('')
This matches anything with a period followed by two or three characters and then a word boundary.
Here's some example outputs:
'example.aus.com' // .aus.com
'example.austin.com' // .austin.com
'example.aus.com/howdy' // .aus.com
'example.co.uk/howdy' // .co.uk
Some people might need something a bit cleverer, but this was enough for me with my particular dataset.
Edit
I've realised there are actually quite a few second-level domains which are longer than 3 characters (and allowed). So, again for simplicity, I just removed the character counting element of my regex:
'example.aus.com'.match(/\.\w*\b/g).join('')
Since TLDs now include things with more than three-characters like .wang and .travel, here's a regex that satisfies these new TLDs:
([^.\s]+\.[^.\s]+)$
Strategy: starting at the end of the string, look for one or more characters that aren't periods or whitespace, followed by a single period, followed by one or more characters that aren't periods or whitespace.
http://regexr.com/3bmb3
With capturing groups you can achieve some magix.
For example, consider the following javascript:
let hostname = 'test.something.else.be';
let domain = hostname.replace(/^.+\.([^\.]+\.[^\.]+)$/, '$1');
document.write(domain);
This will result in a string containing 'else.com'. This is because the regex itself will match the complete string and the capturing group will be mapped to $1. So it replaces the complete string 'test.something.else.com' with '$1' which is actually 'else.com'.
The regex isn't pretty and can probably be made more dynamic with things like {3} for defining how many levels deep you want to look for subdomains, but this is just an illustration.
if you want all specific Top Level Domain name then you can write regular expression like this:
[RegularExpression("^(https?:\\/\\/)?(([\\w]+)?\\.?(\\w+\\.((za|zappos|zara|zero|zip|zippo|zm|zone|zuerich|zw))))\\/?$", ErrorMessage = "Is not a valid fully-qualified URL.")]
You can also put more domain name from this link:
https://www.icann.org/resources/pages/tlds-2012-02-25-en
The following regex matches a domain with root and tld extractions (named capture groups) from a url or domain string:
(?:\w+:\/{2})?(?<cs_domain>(?<cs_domain_sub>(?:[\w\-]+\.)*?)(?<cs_domain_root>[\w\-]+(?<cs_domain_tld>(?:\.\w{2})?(?:\.\w{2,3}|\.xn-+\w+|\.site|\.club))))\|
It's hard to say if it is perfect, but it works on all the test data sets that I have put it against including .club, .xn-1234, .co.uk, and other odd endings. And it does it in 5556 steps against 40k chars of logs, so the efficiency seems reasonable too.
If you need to be more specific:
/\.(?:nl|se|no|es|milru|fr|es|uk|ca|de|jp|au|us|ch|it|io|org|com|net|int|edu|mil|arpa)/
Based on http://www.seobythesea.com/2006/01/googles-most-popular-and-least-popular-top-level-domains/
I'm basically not in the clue about regex but I need a regex statement that will recognise anything after the / in a URL.
Basically, i'm developing a site for someone and a page's URL (Local URL of Course) is say (http://)localhost/sweettemptations/available-sweets. This page is filled with custom post types (It's a WordPress site) which have the URL of (http://)localhost/sweettemptations/sweets/sweet-name.
What I want to do is redirect the URL (http://)localhost/sweettemptations/sweets back to (http://)localhost/sweettemptations/available-sweets which is easy to do, but I also need to redirect any type of sweet back to (http://)localhost/sweettemptations/available-sweets. So say I need to redirect (http://)localhost/sweettemptations/sweets/* back to (http://)localhost/sweettemptations/available-sweets.
If anyone could help by telling me how to write a proper regex statement to match everything after sweets/ in the URL, it would be hugely appreciated.
To do what you ask you need to use groups. In regular expression groups allow you to isolate parts of the whole match.
for example:
input string of: aaaaaaaabbbbcccc
regex: a*(b*)
The parenthesis mark a group in this case it will be group 1 since it is the first in the pattern.
Note: group 0 is implicit and is the complete match.
So the matches in my above case will be:
group 0: aaaaaaaabbbb
group 1: bbbb
In order to achieve what you want with the sweets pattern above, you just need to put a group around the end.
possible solution: /sweets/(.*)
the more precise you are with the pattern before the group the less likely you will have a possible false positive.
If what you really want is to match anything after the last / you can take another approach:
possible other solution: /([^/]*)
The pattern above will find a / with a string of characters that are NOT another / and keep it in group 1. Issue here is that you could match things that do not have sweets in the URL.
Note if you do not mind the / at the beginning then just remove the ( and ) and you do not have to worry about groups.
I like to use http://regexpal.com/ to test my regex.. It will mark in different colors the different matches.
Hope this helps.
I may have misunderstood you requirement in my original post.
if you just want to change any string that matches
(http://)localhost/sweettemptations/sweets/*
into the other one you provided (without adding the part match by your * at the end) I would use a regular expression to match the pattern in the URL but them just blind replace the whole string with the desired one:
(http://)localhost/sweettemptations/available-sweets
So if you want the URL:
http://localhost/sweettemptations/sweets/somethingmore.html
to turn into:
http://localhost/sweettemptations/available-sweets
and not into:
localhost/sweettemptations/available-sweets/somethingmore.html
Then the solution is simpler, no groups required :).
when doing this I would make sure you do not match the "localhost" part. Also I am assuming the (http://) really means an optional http:// in front as (http://) is not a valid protocol prefix.
so if that is what you want then this should match the pattern:
(http://)?[^/]+/sweettemptations/sweets/.*
This regular expression will match the http:// part optionally with a host (be it localhost, an IP or the host name). You could omit the .* at the end if you want.
If that pattern matches just replace the whole URL with the one you want to redirect to.
use this regular expression (?<=://).+
I need to match all valid URLs except:
http://www.w3.org
http://w3.org/foo
http://www.tempuri.org/foo
Generally, all URLs except certain domains.
Here is what I have so far:
https?://([-\w\.]+)+(:\d+)?(/([\w/_\.]*(\?\S+)?)?)?
will match URLs that are close enough to my needs (but in no way all valid URLs!) (thanks, http://snipplr.com/view/2371/regex-regular-expression-to-match-a-url/!)
https?://www\.(?!tempuri|w3)\S*
will match all URLs with www., but not in the tempuri or w3 domain.
And I really want
https?://([-\w\.]+)(?!tempuri|w3)\S*
to work, but afaick, it seems to select all http:// strings.
Gah, I should just do this in something higher up the Chomsky hierarchy!
The following regular expression:
https?://(?!w3|tempuri)([-\w]*\.)(?!w3|tempuri)\S*
only matches the first four lines from the following excerpt:
https://ok1.url.com
http://ok2.url.com
https://not.ok.tempuri.com
http://not-ok.either.w3.com
http://no1.w3.org
http://no2.w3.org
http://tempuri.bla.com
http://no4.tempuri.bla
http://no3.tempuri.org
http://w3.org/foo
http://www.tempuri.org/foo
I know what you're thinking, and the answer is that in order to match the above list and only return the first two lines you'd have to use the following regular expression:
https?://(?!w3|tempuri)([-\w]*\.)(?!w3|tempuri)([-\w]*\.)(?!w3|tempuri)\S*
which, in truth, is nothing more than a slight modification of the first regular expression, where the
(?!w3|tempuri)([-\w]*\.)
part appears twice in a row.
The reason why your regular expression wasn't working was because when you include . inside the ()* then that means it can not only match this. and this.this. but also this.this.th - in other words, it doesn't necessarily end in a dot, so it will force it to end wherever it has to so that the expression matches. Try it out in a regular expression tester and you'll see what I mean.