Develop Reference

Parsing url string from '+' to '%2B'

I have url address where its extension needs to be in ASCII/UTF-8
a='sAE3DSRAfv+HG='
i need to convert above as this:
a='sAE3DSRAfv%2BHG%3D'
I searched but not able to get it.

Please see built-in method urllib.parse.quote()
A very important task for the URL is its safe transmission. Its meaning must not change after you created it till it is received by the intended receiver. To achieve that end URL encoding was incorporated. See RFC 2396
URL might contain non-ascii characters like cafés, López etc. Or it might contain symbols which have different meaning when put in the context of a URL. For example, # which signifies a bookmark. To ensure safe transmitting of such characters HTTP standards maintain that you quote the url at the point of origin. And URL is always present in quoted format to anyone else.
I have put sample usage below.
>>> import urllib.parse
>>> a='sAE3DSRAfv+HG='
>>> urllib.parse.quote(a)
'sAE3DSRAfv%2BHG%3D'
>>>

Python - Remove 'style'-attribute from HTML

I have a String in Python, which has some HTML in it. Basically it looks like this.
>>> print someString # I get someString from the backend
"<img style='height:50px;' src='somepath'/>"
I try to display this HTML in a PDF. Because my PDF generator can't handle the styles-attribute (and no, I can't take another one), I have to remove it from the string. So basically, it should be like that:
>>> print someString # I get someString from the backend
"<img style='height:50px;' src='somepath'/>"
>>> parsedString = someFunction(someString)
>>> print parsedString
"<img src='somepath'/>"
I guess the best way to do this is with RegEx, but I'm not very keen on it. Can someone help me out?

I wouldn't use RegEx with this because
Regex is not really suited for HTML parsing and even though this is a simple one there could be many variations and edge cases you need to consider and the resulting regex could turn out to be a nightmare
Regex sucks. It can be really useful but honestly, they are the epitome of user unfriendly.
Alright, so how would I go about it. I would use trusty BeautifulSoup! Install with pip by using the following command:
pip install beautifulsoup4
Then you can do the following to remove the style:
from bs4 import BeautifulSoup as Soup
del Soup(someString).find('img')['style']
This first parses your string, then finds the img tag and then deletes its style attribute.
It should also work with arbitrary strings but I can't promise that. Maybe you will come up with an edge case.
Remember, using RegEx to parse an HTML string is not the best of ideas. The internet and Stackoverflow is full of answers why this is not possible.
Edit: Just for kicks you might want to check out this answer. You know stuff is serious when it is said that even Jon Skeet can't do it.

Using RegEx to work with HTML is a very bad idea but if you really want to use it, try this:
/style=["']?((?:.(?!["']?\s+(?:\S+)=|[>"']))+.)["']?/ig

Python: How to check if a string is a valid IRI?

Is there a standard function to check an IRI, to check an URL apparently I can use:
parts = urlparse.urlsplit(url)
if not parts.scheme or not parts.netloc:
'''apparently not an url'''
I tried the above with an URL containing Unicode characters:
import urlparse
url = "http://fdasdf.fdsfîășîs.fss/ăîăî"
parts = urlparse.urlsplit(url)
if not parts.scheme or not parts.netloc:
print "not an url"
else:
print "yes an url"
and what I get is yes an url. Does this means I'm good an this tests for valid IRI? Is there another way ?

Using urlparse is not sufficient to test for a valid IRI.
Use the rfc3987 package instead:
from rfc3987 import parse
parse('http://fdasdf.fdsfîășîs.fss/ăîăî', rule='IRI')

The only character-set-sensitive code in the implementation of urlparse is requiring that the scheme should contain only ASCII letters, digits and [+-.] characters; otherwise it's completely agnostic so will work fine with non-ASCII characters.
As this is non-documented behaviour, it's your responsibility to check that it continues to be the case (with tests in your project), but I don't imagine it would be changed to break IRIs.
urllib provides quoting functions to convert IRIs to/from ASCII URIs, although they still don't mention IRIs explicitly in the documentation, and they are broken in some cases: Is there a unicode-ready substitute I can use for urllib.quote and urllib.unquote in Python 2.6.5?

Python unescape URL

I have got a url in this form - http:\\/\\/en.wikipedia.org\\/wiki\\/The_Truman_Show. How can I make it normal url. I have tried using urllib.unquote without much success.
I can always use regular expressions or some simple string replace stuff. But I believe that there is a better way to handle this...

urllib.unquote is for replacing %xx escape codes in URLs with the characters they represent. It won't be useful for this.
Your "simple string replace stuff" is probably the best solution.

Have you tried using json.loads from the json module?
>>> json.loads('"http:\\/\\/en.wikipedia.org\\/wiki\\/The_Truman_Show"')
'http://en.wikipedia.org/wiki/The_Truman_Show'
The input that I'm showing isn't exactly what you have. I've wrapped it in double quotes to make it valid json.
When you first get it from the json, how are you decoding it? That's probably where the problem is.

It is too childish -- look for some library function when you can transform URL by yourself.
Since there are not other visible rules but "/" replaced by "\/", you can simply replace it back:
def unescape_this(url):
return url.replace(r"\\/", "/")

[Python]How to deal with a string ending with one backslash?

I'm getting some content from Twitter API, and I have a little problem, indeed I sometimes get a tweet ending with only one backslash.
More precisely, I'm using simplejson to parse Twitter stream.
How can I escape this backslash ?
From what I have read, such raw string shouldn't exist ...
Even if I add one backslash (with two in fact) I still get an error as I suspected (since I have a odd number of backslashes)
Any idea ?
I can just forget about these tweets too, but I'm still curious about that.
Thanks : )

Prepending the string with r (stands for "raw") will escape all characters inside the string. For example:
print r'\b\n\\'
will output
\b\n\\
Have I understood the question correctly?

I guess you are looking a method similar to stripslashes in PHP. So, here you go:
Python version of PHP's stripslashes

You can try using raw strings by prepending an r (so nothing has to be escaped) to the string or re.escape().
I'm not really sure what you need considering I haven't seen the text of the response. If none of the methods you come up with on your own or get from here work, you may have to forget about those tweets.

Unless you update your question and come back with a real problem, I'm asserting that you don't have an issue except confusion.
You get the string from the Tweeter API, ergo the string does not show up in your code. “Raw strings” exist only in your code, and it is “raw strings” in code that can't end in a backslash.
Consider this:
def some_obscure_api():
"This exists in a library, so you don't know what it does"
return r"hello" + "\\" # addition just for fun
my_string = some_obscure_api()
print(my_string)
See? my_string happily ends in a backslash and your code couldn't care less.