Getting an empty jpg file, when try to Download an image - python

I have been trying to download an image from website (no username and password required) but every time I am getting an empty file. I have used conventional urllib .retrieve and requests methodologies but getting the same result. One thing more is that if I try to open the same image manually by copy pasting the URL after 15-20 min then that image itself does not open. I am assuming that some sort of session handling is required in this case . Below is my code which returns me empty image.
import os
import urllib
def savePic(url):
uri="C:\Python27\Scripts\Photosurl2.jpg"
if url!="":
urllib.urlretrieve(url, uri)
savePic("http://www-nass.nhtsa.dot.gov/nass/cds/GetBinary.aspx?ImageView&ImageID=491410290&Desc=Lookback+from+final+rest&Title=Scene+Photos+-+image1&Version=1&Extend=jpg")
Any help is appreciated.

When you try to implement some HTTP code in Python do not forget to validate that you can use curl or wget to perform these HTTP requests. This will save you a lot of time trying to debug a problem that is not in your code.
The also have very good verbose modes which will give you some hints regarding what you are missing.
Also, most senior Python developers are using the requests library instead of the urllib ones.
PS. Requests library is easier to use than urllib.

Related

Trouble submitting image URL from Reddit using requests.get()

I am trying to submit an image URL from Reddit.com to a vision API using requests.get() in Python but I am running into difficulties in what could be a simple error on my part. The requests.get() request is successful when the link points to an explicit *.jpg, e.g., https://upload.wikimedia.org/wikipedia/commons/thumb/2/2b/Beef_fillet_steak_with_mushrooms.jpg/800px-Beef_fillet_steak_with_mushrooms.jpg, but unsuccessful when the link points to what I perceive to be a soft link, e.g., https://preview.redd.it/9xu97c5snpr51.jpg?width=640&crop=smart&auto=webp&s=e68c02166f6fd21a47a957b187b98b92608f54a9. Note that when pasted into a browser, both links work fine.
Does anyone have a suggestion for how I might preprocess the second link so it is handled like the first link? I would like to eventually have this code run remotely, so avoiding having to download the file locally is preferred.
From the documentation on: https://requests.readthedocs.io/en/master/user/quickstart/
You can access the response body as bytes, for non-text requests:
from PIL import Image
from io import BytesIO
i = Image.open(BytesIO(r.content))

Using Python, upload image on https://cloud.google.com/vision/ and read JSON script

I want to write a python script to automate uploading of image on https://cloud.google.com/vision/ and collect information from JSON tab there. I need to know how to do it.
Till now, I'm only able to open the website on chrome using following code:-
import webbrowser
url = 'https://cloud.google.com/vision/'
webbrowser.open_new_tab(url + 'doc/')
I tried using urllib2 but couldn't get anything.
Help me out please
you have to use google-cloud-vision lib
there is a sample code in this docs
https://cloud.google.com/vision/docs/reference/libraries#client-libraries-install-python
you can start from here

urllib.urlretrieve is failing

I'm trying to download data using commands below.
import urllib
url = 'http://www.nse-india.com/content/historical/EQUITIES/2002/MAR/cm01MAR2002bhav.csv.zip'
urllib.urlretrieve(url, 'myzip')
What I see in the file generated file my.zip is,
You don't have permission to access "http://www.nse-india.com/content/historical/EQUITIES/2002/MAR/cm01MAR2002bhav.csv.zip" on this server.<P>
Reference #18.7d427b5c.1311889977.25329891
But I'm able to download the file from the website without any problem.
What is the reason for this?
You may need to use urllib2 and set the user-agent header to something it recognizes. It might just be blocking anything that doesn't appear to be a normal user.

python downloading data (urllib, urllib2)

I have a link like this, direct to a mp3 file. So when I put it in my browser, basically asks me if I want to download the file, however when I do the same thing with python by the following code :
> data = urllib2.urlopen("http://www23.zippyshare.com/d/44123087/497548/Lil%20Wayne%20ft.%20Eminem%20-%20Drop%20The%20World.mp3".read())
I will redirected to another link like this. Therefore, instead of the MP3 data, I am getting the html code for
'http://www23.zippyshare.com/v/44123087/file.html'
any ideas ?
thanks
urllib2 handles redirection transparently. You might want to see what the server is actually doing when it is presenting such a redirection as well allowing you to download. You might want to subclass the redirect handler and see which property of the header is giving you the url and use urlretrieve to download that.
Setting the cookies, trying explicitly might be a good try as well.
import cookielib, urllib2
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj))
opener.open('yourmp3filelink')
Your link redirects to an HTML webpage, most likely because your download request is timing out. That's often how these download websites work: you never get a static link to the download, only a temporarily assigned link.
My guess is that there's no way to get that static link using that website. You'd have to know where that file was actually coming from.
So no, nothing is wrong with your python code; just your sources.

Using Python to download a document that's not explicitly referenced in a URL

I wrote a web crawler in Python 2.6 using the Bing API that searches for certain documents and then downloads them for classification later. I've been using string methods and urllib.urlretrieve() to download results whose URL ends in .pdf, .ps etc., but I run into trouble when the document is 'hidden' behind a URL like:
http://www.oecd.org/officialdocuments/displaydocument/?cote=STD/CSTAT/WPNA(2008)25&docLanguage=En
So, two questions. Is there a way in general to tell if a URL has a pdf/doc etc. file that it's linking to if it's not doing so explicitly (e.g. www.domain.com/file.pdf)? Is there a way to get Python to snag that file?
Edit:
Thanks for replies, several of which suggest downloading the file to see if it's of the correct type. Only problem is... I don't know how to do that (see question #2, above). urlretrieve(<above url>) gives only an html file with an href containing that same url.
There's no way to tell from the URL what it's going to give you. Even if it ends in .pdf it could still give you HTML or anything it likes.
You could do a HEAD request and look at the content-type, which, if the server isn't lying to you, will tell you if it's a PDF.
Alternatively you can download it and then work out whether what you got is a PDF.
In this case, what you refer to as "a document that's not explicitly referenced in a URL" seems to be what is known as a "redirect". Basically, the server tells you that you have to get the document at another URL. Normally, python's urllib will automatically follow these redirects, so that you end up with the right file. (and - as others have already mentioned - you can check the response's mime-type header to see if it's a pdf).
However, the server in question is doing something strange here. You request the url, and it redirects you to another url. You request the other url, and it redirects you again... to the same url! And again... And again... At some point, urllib decides that this is enough already, and will stop following the redirect, to avoid getting caught in an endless loop.
So how come you are able to get the pdf when you use your browser? Because apparently, the server will only serve the pdf if you have cookies enabled. (why? you have to ask the people responsible for the server...) If you don't have the cookie, it will just keep redirecting you forever.
(check the urllib2 and cookielib modules to get support for cookies, this tutorial might help)
At least, that is what I think is causing the problem. I haven't actually tried doing it with cookies yet. It could also be that the server is does not "want" to serve the pdf, because it detects you are not using a "normal" browser (in which case you would probably need to fiddle with the User-Agent header), but it would be a strange way of doing that. So my guess is that it is somewhere using a "session cookie", and in the case you haven't got one yet, keeps on trying to redirect.
As has been said there is no way to tell content type from URL. But if you don't mind getting the headers for every URL you can do this:
obj = urllib.urlopen(URL)
headers = obj.info()
if headers['Content-Type'].find('pdf') != -1:
# we have pdf file, download whole
...
This way you won't have to download each URL just it's headers. It's still not exactly saving network traffic, but you won't get better than that.
Also you should use mime-types instead of my crude find('pdf').
No. It is impossible to tell what kind of resource is referenced by a URL just by looking at it. It is totally up to the server to decide what he gives you when you request a certain URL.
Check the mimetype with the urllib.info() function. This might not be 100% accurate, it really depends on what the site returns as a Content-Type header. If it's well behaved it'll return the proper mime type.
A PDF should return application/pdf, but that may not be the case.
Otherwise you might just have to download it and try it.
You can't see it from the url directly. You could try to only download the header of the HTTP response and look for the Content-Type header. However, you have to trust the server on this - it could respond with a wrong Content-Type header not matching the data provided in the body.
Detect the file type in Python 3.x and webapp with url to the file which couldn't have an extension or a fake extension. You should install python-magic, using
pip3 install python-magic
For Mac OS X, you should also install libmagic using
brew install libmagic
Code snippet
import urllib
import magic
from urllib.request import urlopen
url = "http://...url to the file ..."
request = urllib.request.Request(url)
response = urlopen(request)
mime_type = magic.from_buffer(response.read())
print(mime_type)

Categories