How to use exceptions for different cases with python requests - python

I have this code
try:
response = requests.post(url, data=json.dumps(payload))
except (ConnectionError, HTTPError):
msg = "Connection problem"
raise Exception(msg)
Now i want the following
if status_code == 401
login() and then try request again
if status_code == 400
then send respose as normal
if status_code == 500
Then server problem , try the request again and if not successful raise EXception
Now these are status codes , i donn't know how can i mix status codes with exceptions. I also don't know what codes will be covered under HttpError


requests has a call called raise_for_status available in your request object which will raise an HTTPError exception if any code is returned in the 400 to 500 range inclusive.
Documentation for raise_for_status is here
So, what you can do, is after you make your call:
response = requests.post(url, data=json.dumps(payload))
You make a call for raise_for_status as
response.raise_for_status()
Now, you are already catching this exception, which is great, so all you have to do is check to see which status code you have in your error. This is available to you in two ways. You can get it from your exception object, or from the request object. Here is the example for this:
from requests import get
from requests.exceptions import HTTPError
try:
r = get('http://google.com/asdf')
r.raise_for_status()
except HTTPError as e:
# Get your code from the exception object like this
print(e.response.status_code)
# Or you can get the code which will be available from r.status_code
print(r.status_code)
So, with the above in mind, you can now use the status codes in your conditional statements


https://docs.python.org/2/library/urllib2.html#urllib2.URLError
code
An HTTP status code as defined in RFC 2616. This numeric value
corresponds to a value found in the dictionary of codes as found in
BaseHTTPServer.BaseHTTPRequestHandler.responses.
You can get the error code from an HTTPError from its code member, like so
try:
# ...
except HTTPError as ex:
status_code = ex.code

Related

How to test try/except block with pytest and cover all exceptions

I have a function call an api to get an access token :
def get_token():
try:
url = 'https://call-api/accesstoken'
token = 'authentication-token'
headers = {"Authorization": "Bearer %s" % token}
session = requests.Session()
response = session.get(url, headers=headers)
return response.json()
except HTTPError as http_err:
print(f'HTTP error occurred: {http_err}')
except Exception as err:
print(f'Other error occurred: {err}')
I want to test this function with pytest and verify the try/except block.
Here is my test function :
import responses
import functions as m_functions
#responses.activate
def test_get_token():
url = 'https://call-api/accesstoken'
responses.add(responses.GET, url,
json = {
"accessToken": "tmp-token1"
})
response = m_functions.get_token()
assert response['accessToken'] == 'tmp-token1'
The part inside the block try is tested, but how to test the exceptions part ?
For the moment, only part inside try block is covered by tests. I would like my code to be fully covered.
Any idea? Thanks

As shown in the documentation, you can specify an exception as the response body for the request to trigger an exception:
responses.add(responses.GET, url, body=SomeExceptionObject(...))
Therefore, you can make two tests: one where you specify HTTPError as the exception and another tests with a different exception. You can then use the capsys fixture to capture the standard output.

python requests library .json() + django 500

When I use the requests library with django and I get a 500 error back. response.json() gives me this error:
response = requests.post(....)
print("-----------------------block found!!!-----------------------")
print(response.status_code)
print(response.json())
json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char
0)
Is there a way to represent a django 500 response with the requests library in a readable manner?

Assuming you get an HTTP 500 response, You would definitely receive an empty source, Which prevents the .json() function from taking place.
Why not write an exception clause to handle the exceptions, like below:
try:
response = requests.post(....)
print("-----------------------block found!!!-----------------------")
print(response.status_code)
print(response.json())
except HTTPError as e:
print('Error has occurred: ', e.response.status_code)

How get status code if an error occurred python requests?

I want to get response status code if an error occurred
try:
r = requests.post('site.com')
except (requests.RequestException) as e:
print(r.status_code)
i can't do that
I also know about
r.raise_for_status()
But Exception raised before request so i can't do that

Python check if website exists

I wanted to check if a certain website exists, this is what I'm doing:
user_agent = 'Mozilla/20.0.1 (compatible; MSIE 5.5; Windows NT)'
headers = { 'User-Agent':user_agent }
link = "http://www.abc.com"
req = urllib2.Request(link, headers = headers)
page = urllib2.urlopen(req).read() - ERROR 402 generated here!
If the page doesn't exist (error 402, or whatever other errors), what can I do in the page = ... line to make sure that the page I'm reading does exit?

You can use HEAD request instead of GET. It will only download the header, but not the content. Then you can check the response status from the headers.
For python 2.7.x, you can use httplib:
import httplib
c = httplib.HTTPConnection('www.example.com')
c.request("HEAD", '')
if c.getresponse().status == 200:
print('web site exists')
or urllib2:
import urllib2
try:
urllib2.urlopen('http://www.example.com/some_page')
except urllib2.HTTPError, e:
print(e.code)
except urllib2.URLError, e:
print(e.args)
or for 2.7 and 3.x, you can install requests
import requests
response = requests.get('http://www.example.com')
if response.status_code == 200:
print('Web site exists')
else:
print('Web site does not exist')

It's better to check that status code is < 400, like it was done here. Here is what do status codes mean (taken from wikipedia):
1xx - informational
2xx - success
3xx - redirection
4xx - client error
5xx - server error
If you want to check if page exists and don't want to download the whole page, you should use Head Request:
import httplib2
h = httplib2.Http()
resp = h.request("http://www.google.com", 'HEAD')
assert int(resp[0]['status']) < 400
taken from this answer.
If you want to download the whole page, just make a normal request and check the status code. Example using requests:
import requests
response = requests.get('http://google.com')
assert response.status_code < 400
See also similar topics:
Python script to see if a web page exists without downloading the whole page?
Checking whether a link is dead or not using Python without downloading the webpage
How do you send a HEAD HTTP request in Python 2?
Making HTTP HEAD request with urllib2 from Python 2

from urllib2 import Request, urlopen, HTTPError, URLError
user_agent = 'Mozilla/20.0.1 (compatible; MSIE 5.5; Windows NT)'
headers = { 'User-Agent':user_agent }
link = "http://www.abc.com/"
req = Request(link, headers = headers)
try:
page_open = urlopen(req)
except HTTPError, e:
print e.code
except URLError, e:
print e.reason
else:
print 'ok'
To answer the comment of unutbu:
Because the default handlers handle redirects (codes in the 300 range), and codes in the 100-299 range indicate success, you will usually only see error codes in the 400-599 range.
Source

There is an excellent answer provided by #Adem Öztaş, for use with httplib and urllib2. For requests, if the question is strictly about resource existence, then the answer can be improved upon in the case of large resource existence.
The previous answer for requests suggested something like the following:
def uri_exists_get(uri: str) -> bool:
try:
response = requests.get(uri)
try:
response.raise_for_status()
return True
except requests.exceptions.HTTPError:
return False
except requests.exceptions.ConnectionError:
return False
requests.get attempts to pull the entire resource at once, so for large media files, the above snippet would attempt to pull the entire media into memory. To solve this, we can stream the response.
def uri_exists_stream(uri: str) -> bool:
try:
with requests.get(uri, stream=True) as response:
try:
response.raise_for_status()
return True
except requests.exceptions.HTTPError:
return False
except requests.exceptions.ConnectionError:
return False
I ran the above snippets with timers attached against two web resources:
1) http://bbb3d.renderfarming.net/download.html, a very light html page
2) http://distribution.bbb3d.renderfarming.net/video/mp4/bbb_sunflower_1080p_30fps_normal.mp4, a decently sized video file
Timing results below:
uri_exists_get("http://bbb3d.renderfarming.net/download.html")
# Completed in: 0:00:00.611239
uri_exists_stream("http://bbb3d.renderfarming.net/download.html")
# Completed in: 0:00:00.000007
uri_exists_get("http://distribution.bbb3d.renderfarming.net/video/mp4/bbb_sunflower_1080p_30fps_normal.mp4")
# Completed in: 0:01:12.813224
uri_exists_stream("http://distribution.bbb3d.renderfarming.net/video/mp4/bbb_sunflower_1080p_30fps_normal.mp4")
# Completed in: 0:00:00.000007
As a last note: this function also works in the case that the resource host doesn't exist. For example "http://abcdefghblahblah.com/test.mp4" will return False.

I see many answers that use requests.get, but I suggest you this solution using only requests.head which is faster and also better for the webserver since it doesn't need to send back the body too.
import requests
def check_url_exists(url: str):
"""
Checks if a url exists
:param url: url to check
:return: True if the url exists, false otherwise.
"""
return requests.head(url, allow_redirects=True).status_code == 200
The meta-information contained in the HTTP headers in response to a HEAD request should be identical to the information sent in response to a GET request.

code:
a="http://www.example.com"
try:
print urllib.urlopen(a)
except:
print a+" site does not exist"

You can simply use stream method to not download the full file. As in latest Python3 you won't get urllib2. It's best to use proven request method. This simple function will solve your problem.
def uri_exists(url):
r = requests.get(url, stream=True)
if r.status_code == 200:
return True
else:
return False

def isok(mypath):
try:
thepage = urllib.request.urlopen(mypath)
except HTTPError as e:
return 0
except URLError as e:
return 0
else:
return 1

Try this one::
import urllib2
website='https://www.allyourmusic.com'
try:
response = urllib2.urlopen(website)
if response.code==200:
print("site exists!")
else:
print("site doesn't exists!")
except urllib2.HTTPError, e:
print(e.code)
except urllib2.URLError, e:
print(e.args)

requests.HTTPError uncaught after a requests.get() 404 response

I'm having a slight problem with the requests library.
Say for example I have a statement like this in Python:
try:
request = requests.get('google.com/admin') #Should return 404
except requests.HTTPError, e:
print 'HTTP ERROR %s occured' % e.code
For some reason the exception is not being caught. I've checked the API documentation for requests but it's a bit slim. Is there anyone who has more experience with the library that might be able to help me out?

Interpreter is your friend:
import requests
requests.get('google.com/admin')
# MissingSchema: Invalid URL u'google.com/admin': No schema supplied
Also, requests exceptions:
import requests.exceptions
dir(requests.exceptions)
Also notice that by default requests doesn't raise exception if status is not 200:
In [9]: requests.get('https://google.com/admin')
Out[9]: <Response [503]>
There is raise_for_status() method that does it:
In [10]: resp = requests.get('https://google.com/admin')
In [11]: resp
Out[11]: <Response [503]>
In [12]: resp.raise_for_status()
...
HTTPError: 503 Server Error: Service Unavailable

Running your code in python 2.7.5:
import requests
try:
response = requests.get('google.com/admin') #Should return 404
except requests.HTTPError, e:
print 'HTTP ERROR %s occured' % e.code
print e
Results in:
File "C:\Python27\lib\site-packages\requests\models.py", line 291, in prepare_url
raise MissingSchema("Invalid URL %r: No schema supplied" % url)
requests.exceptions.MissingSchema: Invalid URL u'google.com/admin': No schema supplied
To get your code to pick up this exception you need to add:
except (requests.exceptions.MissingSchema) as e:
print 'Missing schema occured. status'
print e
Note also it is not a missing schema but a missing scheme.

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