Django: how can I generate reverse urls including numbers in the string? - django

I´m using the following code to generate urls:
url(r'^productos/(?P<person_id>\D+)/$', views.ProductoView, name="producto"),
It works fine with strings that only have letters (dTape).
When I try a string that includes a number (d3Tape) I get a nonreverse error.
Any clues? Maybe my regex is not ok but I can´t find the solution.
Thanks!


You can include all characters by using a character group that includes both \d and \D:
url(r'^productos/(?P<person_id>[\d\D]+)/$', views.ProductoView, name="producto"),
But I'm not sure if this is a good idea. It might be useful to restrict the caracters to a group, like:
url(r'^productos/(?P<person_id>[A-Za-z0-9_-]+)/$', views.ProductoView, name="producto"),
this will include all ASCII alphanumerical characters and a hyphen and underscore.
In django-2.x, I would advise to use the str, or slug path converters:
path('productos/<str:person_id>/', views.ProductoView, name="producto"),

Related

Regex Adding a URL path except the current one I'm at

I'm trying to add something along the lines of this regex logic.
For Input:
reading/
reading/123
reading/456
reading/789
I want the regex to match only
reading/123
reading/456
reading/789
Excluding reading/.
I've tried reading\/* but that doesn't work because it includes reading/

You must escape your backslashes in Hugo, \\/\\d+.

Using RegEx to extract a string in a URL

I've tried to search for this and I'm sure versions of this question have been asked, but I haven't been able to apply other answers to my case.
I need to use RegEx to extract a random string of characters and symbols that appears in the URL when an advertiser sends traffic to me.
The referring URL looks something like this, with the part I want to extract in bold:
https://adclick.g.doubleclick.net/pcs/click%**long-string-of-characters-and-symbols**https://www.mywebsite.com
That long string of characters and symbols (the hash) contains multiple % signs so I need the entire string after the first % sign, but before my website's URL.
I've been pulling my hair out on this and any help would be appreciated!

You can use:
(?<=%).*(?=https)
How it works:
(?<=%) Positive lookbehind: search for a character preceeded by %
.* matches everything until...
(?=https): the first https occurs (Positive lookhead)

How to find "complicated" URLs in a text file

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)

RegEx to cut out URL

I try to get an URL from a String of the following format:
RANDOMRUBBISHhttps://www.my-url.com/randomfirstname_randomlastnameRANDOMRUBBISH
I already tried some things, especially the the look before/after, which I used before successfully on another url format (starts https... ends .html, this was working).
But seems I'm too stupid to figure out the regex for the kind of string mentioned above. I just want the URL part from https.... to the end of the random last name. Is this even possible?
Any Ideas?

If you can guarantee that randomfirstname_randomlastname is all lowercase and RANDOMRUBBISH is all uppercase, you can use character classes [a-z] and [A-Z]. The language the regex is for will determine how to use these.
This is example works in javascript:
var str = "RANDOMRUBBISHhttps://www.my-url.com/randomfirstname_randomlastnameRANDOMRUBBISH";
var match = /https:\/\/www\.my-url\.com\/[a-z]*/.exec(str);

django url matching non characters and charaters

suppose I have this url
url(r'^delete_group/(\w+)/', 'delete_group_view',name='delete_group')
In template
{%url 'delete_group' 'mwas'%} works but when I use
{%url 'delete_group' 'mwas 45'%} is not working. Any way to modify the url to accept both mwas and mwas 45

The issue might be your regex. The URL example you're showing has a space in it. \w won't match spaces. Try this instead: r'^delete_group/([\w\s]+)/ which allows either words or spaces in multiples.
However, know that spaces are not valid in URLs and will likely get converted to %20 or something similar. A best practice is to use hyphens where you would put a space.
I'd also point you at this answer to a similar question.