I need to extract pieces from a URL and I am trying to learn preg_match_all().
The output in a PHP variable:
$content = 'http://www.domain.com/folder1/firstname_lastname.jpg';
Here is my attempt:
preg_match_all('/http://(.*?).jpg/s', $content, $out, PREG_SET_ORDER);
echo $out[0][0] . "\n";
Matching the URL is not easy.
I need to pick out from:
hxxp://www.domain.com/folder1/firstname_lastname.jpg
the following: "www.domain.com" and "folder1" and "firstname_lastname"
Could I get one preg_match_all() for each example?
Thanks in advance.
I like to learn by example and trial and error.
He he... :)
You can use these:
http://uk3.php.net/parse_url
http://uk1.php.net/manual/en/function.parse-str.php
To get constituent parts of a URL and query string.
Related
In LibreOffice, It is possible to run python scripts like this:
sURL = "vnd.sun.star.script:file.function?language=Python&location=document"
oScript = scriptProv.getScript(sURL)
x = oScript.Invoke(args, Array(), Array())
In that example 'file' is a filename, and 'function' is the name of a function in that file.
Is it possible to embed script in that URL? sURL="vnd.." & scriptblock & "?language.."
(It seems like the kind of thing that might be possible with the correct URL, or might not be possible if just not supported).
We can use Python's eval() function. Here is an example inspired by JohnSUN's explanation in the discussion. Note: xray() uses XrayTool to show output, but you could replace that line with any output method of your choosing, such as writing to a file.
def runArbitraryCode(*args):
url = args[0]
codeString = url.split("&codeToRun=")[1]
x = eval(codeString)
xray(x)
Now enter this formula in Calc and Ctrl+click on it.
=HYPERLINK("vnd.sun.star.script:misc_examples.py$runArbitraryCode?language=Python&location=user&codeToRun=5+1")
Result: 6
Obligatory caveat: Running eval() on an unknown string is about the worst idea imaginable in terms of security. So hopefully you're the one controlling the URL and not some black hat hacker!
Let's say I have the URL:
http://host/path/to/../some/thing
I want to parse the URL and get the raw/untouched path string:
/path/to/../some/thing
Using, url::Url, I can do:
let url = url::Url::parse("http://host/path/to/../some/thing").unwrap();
println!("{}", url.path());
But this gives a normalized version of the path:
/path/some/thing
url::Url is doing more than I want it to do. (It also percent encodes things.)
Some other langs make it easy, for example in PHP, I can just do:
$ php -r 'echo parse_url( "http://host/path/to/../some/thing" )["path"] . "\n";'
/path/to/../some/thing
So, can anyone point me towards another URL parsing crate that can do this, and is known to be solid/tested?
I'm looking for a script (or if there isn't, I guess I'll have to write my own).
I wanted to ask if anyone here knows a script that can take a txt file with n links (lets say 200). I need to extract only links that have particular characters in them, let's say I only need links that contain "/r/learnprogramming". I need the script to get those links and write them to another txt files.
Edit: Here is what helped me: grep -i "/r/learnprogramming" 1.txt >2.txt
you can use ajax to read .txt file using jquery
<script src=https://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js></script>
<script>
jQuery(function($) {
console.log("start")
$.get("https://ayulayol.imfast.io/ajaxads/ajaxads.txt", function(wholeTextFile) {
var lines = wholeTextFile.split(/\n/),
randomIndex = Math.floor(Math.random() * lines.length),
randomLine = lines[randomIndex];
console.log(randomIndex, randomLine)
$("#ajax").html(randomLine.replace(/#/g,"<br>"))
})
})
</script>
<div id=ajax></div>
If you are using linux or macOS you could use cat and grep to output the links.
cat in.txt | grep /r/programming > out.txt
Solution provided by OP:
grep -i "/r/learnprogramming" 1.txt >2.txt
Since you did not provide the exact format of the document I assume those links are separated by newline characters. In this case, the code is pretty straightforward using Python/awk since you can iterate over file.readlines() and print only those that match your pattern (either by using a lines.contains(pattern) or using a regex if the pattern is more complex). To store the links in a new file simply redirect the stdout to a new file like this:
python script.py > links.txt
The solution above works even if links are separated by an arbitrary symbol s, first read the file into a single string and split it over s. I hope this helps.
I am trying to detect the urls from a text and replace them by wrapping in quotes like below:
original text: Hey, it is a url here www.example.com
required text: Hey, it is a url here "www.example.com"
original text show my input value and required text represents the required output. I searched a lot on web but could not find any possible solution. I already have tried URL.extract feature but that doesn't seem to detect URLs without http or https. Below are the examples of some of urls I want to deal with. Kindly let me know if you know the solution.
ANQUETIL-DUPERRON Abraham-Hyacinthe, KIEFFER Jean-Luc, www.hominides.net/html/actualites/outils-preuve-presence-hominides-asie-0422.php,Les Belles lettres, 2001.
https://www.ancient-code.com/indian-archeologists-stumbleacross-ruins-great-forgotten-civilization-mizoram/
www.jstor.org/stable/24084454
www.biorespire.com/2016/03/22/une-nouvelle-villeantique-d%C3%A9couverte-en-inde/
insu.cnrs.fr/terre-solide/terre-et-vie/de-nouvellesdatations-repoussent-l-age-de-l-apparition-d-outils-surle-so
www.cerege.fr/spip.php?page=pageperso&id_user=94
Find words who look like urls:
str = "ANQUETIL-DUPERRON Abraham-Hyacinthe, KIEFFER Jean-Luc, www.hominides.net/html/actualites/outils-preuve-presence-hominides-asie-0422.php,Les Belles lettres, 2001.\n\nhttps://www.ancient-code.com/indian-archeologists-stumbleacross-ruins-great-forgotten-civilization-mizoram/\n\nwww.jstor.org/stable/24084454\n\nwww.biorespire.com/2016/03/22/une-nouvelle-villeantique-d%C3%A9couverte-en-inde/\n\ninsu.cnrs.fr/terre-solide/terre-et-vie/de-nouvellesdatations-repoussent-l-age-de-l-apparition-d-outils-surle-so\n\nwww.cerege.fr/spip.php?page=pageperso&id_user=94"
str.split.select{|w| w[/(\b+\.\w+)/]}
This will give you an array of words which have no spaces and include a one or more . characters which MIGHT work for your use case.
puts str.split.select{|w| w[/(\b+\.\w+)/]}
www.hominides.net/html/actualites/outils-preuve-presence-hominides-asie-0422.php,
https://www.ancient-code.com/indian-archeologists-stumbleacross-ruins-great-forgotten-civilization-mizoram/
www.jstor.org/stable/24084454
www.biorespire.com/2016/03/22/une-nouvelle-villeantique-d%C3%A9couverte-en-inde/
insu.cnrs.fr/terre-solide/terre-et-vie/de-nouvellesdatations-repoussent-l-age-de-l-apparition-d-outils-surle-so
www.cerege.fr/spip.php?page=pageperso&id_user=94
Updated
Complete solution to modify your string:
str_with_quote = str.clone # make a clone for the `gsub!`
str.split.select{|w| w[/(\b+\.\w+)/]}
.each{|url| str_with_quote.gsub!(url, '"' + url + '"')}
Now your cloned object wraps urls inside double quotes
puts str_with_quote
Will give you this output
ANQUETIL-DUPERRON Abraham-Hyacinthe, KIEFFER Jean-Luc, "www.hominides.net/html/actualites/outils-preuve-presence-hominides-asie-0422.php,Les" Belles lettres, 2001.
"https://www.ancient-code.com/indian-archeologists-stumbleacross-ruins-great-forgotten-civilization-mizoram/"
"www.jstor.org/stable/24084454"
"www.biorespire.com/2016/03/22/une-nouvelle-villeantique-d%C3%A9couverte-en-inde/"
"insu.cnrs.fr/terre-solide/terre-et-vie/de-nouvellesdatations-repoussent-l-age-de-l-apparition-d-outils-surle-so"
"www.cerege.fr/spip.php?page=pageperso&id_user=94"
Suppose I want to turn this :
http://en.wikipedia.org/wiki/Anarchy
into this :
en.wikipedia.org
or even better, this :
wikipedia.org
Is this even possible in regex?
Why use a regex when Ruby has a library for it? The URI library:
ruby-1.9.1-p378 > require 'uri'
=> true
ruby-1.9.1-p378 > uri = URI.parse("http://en.wikipedia.org/wiki/Anarchy")
=> #<URI::HTTP:0x000001010a2270 URL:http://en.wikipedia.org/wiki/Anarchy>
ruby-1.9.1-p378 > uri.host
=> "en.wikipedia.org"
ruby-1.9.1-p378 > uri.host.split('.')
=> ["en", "wikipedia", "org"]
Splitting the host is one way to separate the domains, but I'm not aware of a reliable way to get the base domain -- you can't just count, in the event of a URL like "http://somedomain.otherdomain.school.ac.uk" vs "www.google.com".
/http:\/\/([^\/]*).*/ will produce en.wikipedia.org from the string you provided.
/http:\/\/.{0,3}\.([^\/]*).*/ will produce wikipedia.org.
yes
Now I know you haven't asked for how, and you haven't specified a language, but I'll answer anyway... (note, this works for all language subsites, not just en.wikipedia...)
perl:
$url =~ s,http://[a-z]{2}\.(wikipedia\.org)/.*,$1,;
ruby:
url = url.sub(/http:\/\/[a-z]{2}\.(wikipedia\.org)\/.*/, '\1')
php:
$url = preg_replace('|http://[a-z]{2}.(wikipedia.org)/.*|, '$1', $url);
Of course, for this particular example, you don't even need a regex, just this will do:
url = 'wikipedia.org'
but I jest...
you probably want to handle any URL and pull out the domain part, and it should also work for domains in different countries, eg: foo.co.uk.
In which case, I'd use Mark Rushakoff's solution to get the hostname and then a regex to pull out the domain:
domain = host.sub(/^.*\.([^.]+\.[^.]+(\.[a-z]{2})?)$/, '\1')
Hope this helps
Also, if you want to learn more, I have a regex tute online: http://tech.bluesmoon.info/2006/04/beginning-regular-expressions.html
Sure all you would have to do is search on http://(.*)/wiki/Anarchy
In Perl (Sorry I don't know Ruby, but I expect it's similar)
$string_to_search =~ s/http:////(.)//. should give you wikipedia.org
to get rid of the en, you can simply search on http:////en(.)//......
That should do it.
Update: In case you're not familiar with Regex, I would recommend picking up a Regex book, this one really rocks and I like it: REGEX BOOK,Mastering Regular Expressions, I saw it on half.com the other day for 14.99 used, but to clarify what i suggested above, is to look for the string http://en, then for anything until you find a / this is all captured in $1 (in perl, not sure if it's the same in ruby), a simple print $1 will print the string.
Update: #2 sorry the star in the regex is not showing up for some reason, so where you see the . in the () and after the // just imagine a *, oh and I forgot for the en part add a /. at the end that way you don't end up with .wikipedia.org