I would like to be able to enter a server response code and have Requests tell me what the code means. For example, code 200 --> ok
I found a link to the source code which shows the dictionary structure of the codes and descriptions. I see that Requests will return a response code for a given description:
print requests.codes.processing # returns 102
print requests.codes.ok # returns 200
print requests.codes.not_found # returns 404
But not the other way around:
print requests.codes[200] # returns None
print requests.codes.viewkeys() # returns dict_keys([])
print requests.codes.keys() # returns []
I thought this would be a routine task, but cannot seem to find an answer to this in online searching, or in the documentation.
Alternatively, in case of Python 2.x, you can use httplib.responses:
>>> import httplib
>>> httplib.responses[200]
'OK'
>>> httplib.responses[404]
'Not Found'
In Python 3.x, use http module:
In [1]: from http.client import responses
In [2]: responses[200]
Out[2]: 'OK'
In [3]: responses[404]
Out[3]: 'Not Found'
One possibility:
>>> import requests
>>> requests.status_codes._codes[200]
('ok', 'okay', 'all_ok', 'all_okay', 'all_good', '\\o/', '\xe2\x9c\x93')
The first value in the tuple is used as the conventional code key.
I had the same problem before and found the
answer in this question
Basically:
responsedata.status_code - gives you the integer status code
responsedata.reason - gives the text/string representation of the status code
requests.status_codes.codes.OK
works nicely and makes it more readable in my application code
Notice that in source code: the requests.status_codes.codes is of type LookupDict which overrides method getitem
You could see all the supported keys with - dir(requests.status_codes.codes)
When using in combination with FLASK:
i like use following enum from flask-api plugin
from flask_api import status where i get more descriptive version of HTTP status codes as in -
status.HTTP_200_OK
With Python 3.x this will work
>>> from http import HTTPStatus
>>> HTTPStatus(200).phrase
'OK'
Related
I'm new to Python, and I'm working in Pycharm to read data line by line from a webpage. For this task, I'm attempting to use the requests module. However, when I try to print the response object, I see "Process finished with exit code 0" and no object displayed.
Do I need to create some sort of setting to be able to work with HTTP requests in Python?
Code:
import re
import requests
def find_phone_number(url='https://www.python-course.eu/barneyhouse.txt'):
response = requests.get(url)
return response
print(find_phone_number(url='https://www.python-course.eu/barneyhouse.txt'))
You need to call the function and access the 'text' element.
Also, in your code the print statement is not indented properly so it will never be run.
Here is an example of the code doing what I think you intendend:
import re
import requests
def find_phone_number(url='https://www.python-course.eu/simpsons_phone_book.txt'):
response = requests.get(url)
return response
text_you_want = find_phone_number().text
print(text_you_want)
Well, for starters, your find_phone_number() function calls itself after it returns. This is because your last line is indented and therefore inside the function definition. The reason you keep getting Process finished with exit code 0 is because your function is never actually called. This should work:
import re
import requests
def find_phone_number(url='https://www.python-course.eu/barneyhouse.txt'):
response = requests.get(url)
return response
print(find_phone_number(url='https://www.python-course.eu/barneyhouse.txt'))
Here is the link im opening in python:
response = urllib.request.urlopen('http://freegeoip.net/json/1.2.3.4').read()
print(response)
After printing the response it gives the result:
b'{"ip":"1.2.3.4","country_code":"US","country_name":"United States","region_code":"WA","region_name":"Washington","city":"Mukilteo","zip_code":"98275","time_zone":"America/Los_Angeles","latitude":47.913,"longitude":-122.305,"metro_code":819}\n'
My question is, how can i print just the region name? i have this code so far:
import json
import urllib.request
response = urllib.request.urlopen('http://freegeoip.net/json/1.2.3.4').read()
print(response)
result = json.loads(response.decode('utf8'))
Its the last bit of pulling out the specific piece of data im stuck on. Thanks in advance!
Your result object returned from json.loads will be a dictionary, so you can print the value like so:
print result['region_name']
Or even better, a bit more defensively, in the event that key doesn't exist:
print result.get('region_name', 'No region specified')
Also note that your urllib call should be:
response = urllib.urlopen('http://freegeoip.net/json/1.2.3.4').read()
At this point, you'd be able to access it as you would a python object:
result['region_code']
you can use ast.literal_eval:
>>> import ast
>>> my_dict = ast.literal_eval(response.strip())
>>> my_dict['region_name']
'Washington'
I need to perform http PUT operations from python Which libraries have been proven to support this? More specifically I need to perform PUT on keypairs, not file upload.
I have been trying to work with the restful_lib.py, but I get invalid results from the API that I am testing. (I know the results are invalid because I can fire off the same request with curl from the command line and it works.)
After attending Pycon 2011 I came away with the impression that pycurl might be my solution, so I have been trying to implement that. I have two issues here. First, pycurl renames "PUT" as "UPLOAD" which seems to imply that it is focused on file uploads rather than key pairs. Second when I try to use it I never seem to get a return from the .perform() step.
Here is my current code:
import pycurl
import urllib
url='https://xxxxxx.com/xxx-rest'
UAM=pycurl.Curl()
def on_receive(data):
print data
arglist= [\
('username', 'testEmailAdd#test.com'),\
('email', 'testEmailAdd#test.com'),\
('username','testUserName'),\
('givenName','testFirstName'),\
('surname','testLastName')]
encodedarg=urllib.urlencode(arglist)
path2= url+"/user/"+"99b47002-56e5-4fe2-9802-9a760c9fb966"
UAM.setopt(pycurl.URL, path2)
UAM.setopt(pycurl.POSTFIELDS, encodedarg)
UAM.setopt(pycurl.SSL_VERIFYPEER, 0)
UAM.setopt(pycurl.UPLOAD, 1) #Set to "PUT"
UAM.setopt(pycurl.CONNECTTIMEOUT, 1)
UAM.setopt(pycurl.TIMEOUT, 2)
UAM.setopt(pycurl.WRITEFUNCTION, on_receive)
print "about to perform"
print UAM.perform()
httplib should manage.
http://docs.python.org/library/httplib.html
There's an example on this page http://effbot.org/librarybook/httplib.htm
urllib and urllib2 are also suggested.
Thank you all for your assistance. I think I have found an answer.
My code now looks like this:
import urllib
import httplib
import lxml
from lxml import etree
url='xxxx.com'
UAM=httplib.HTTPSConnection(url)
arglist= [\
('username', 'testEmailAdd#test.com'),\
('email', 'testEmailAdd#test.com'),\
('username','testUserName'),\
('givenName','testFirstName'),\
('surname','testLastName')\
]
encodedarg=urllib.urlencode(arglist)
uuid="99b47002-56e5-4fe2-9802-9a760c9fb966"
path= "/uam-rest/user/"+uuid
UAM.putrequest("PUT", path)
UAM.putheader('content-type','application/x-www-form-urlencoded')
UAM.putheader('accepts','application/com.internap.ca.uam.ama-v1+xml')
UAM.putheader("Content-Length", str(len(encodedarg)))
UAM.endheaders()
UAM.send(encodedarg)
response = UAM.getresponse()
html = etree.HTML(response.read())
result = etree.tostring(html, pretty_print=True, method="html")
print result
Updated: Now I am getting valid responses. This seems to be my solution. (The pretty print at the end isn't working, but I don't really care, that is just there while I am building the function.)
>>> import httplib
>>> conn = httplib.HTTPConnection("www.google.com")
>>> conn.request("HEAD", "/index.html")
>>> res = conn.getresponse()
>>> print res.status, res.reason
200 OK
This code will get the HTTP status code. However, notice that I split up "google.com" and "/index.html" on 2 lines.
And it's confusing.
What if I want to find the status code of just a general URL???
http://mydomain.com/sunny/boo.avi
http://anotherdomain.com/podcast.mp3
http://anotherdomain.com/rss/fee.xml
Can't I just stick the URL into it, and make it work?
Edit...I cannot use urllib, because I don't want to downlaod the file
Alternatively, if you expect that actually downloading the data is problematic and you really need the HEAD method, you could parse the URL using urlparse:
>>> import httplib
>>> import urlparse
>>> url = "http://www.google.com/index.html"
>>> (scheme, netloc, path, params, query, fragment) = urlparse.urlparse(url)
>>> conn = httplib.HTTPConnection(netloc)
>>> conn.request("HEAD", urlparse.urlunparse(('', '', path, params, query, fragment))
>>> res = conn.getresponse()
>>> print res.status, res.reason
302 Found
And wrap this into a function taking the URL as an argument.
Maybe you are better off using the URL library instead?
In Python 2, use urllib2:
>>> import urllib2
>>> url = urllib2.urlopen("http://www.google.com/index.html")
>>> url.getcode()
200
In Python 3, use urllib.request:
>>> import urllib.request
>>> url = urllib.request.urlopen("http://www.google.com/index.html")
>>> url.getcode()
200
The connect method takes a server argument (with an optional port). You have to split the connection with the resource you actually want.
For a simpler way to download web resources directly, you could go with urllib2 but urllib2 only supports GET or POST methods, no HEAD, so you end up downloading the whole resource.
According to the spec you're supposed to split it up like that, maybe Python could abstract that out for you a bit, they're probably just giving you straight access to the header so you know exactly how it's being formatted, which is really the preference.
Keep in mind that not all web servers support HEAD on each resource so you'll end up retrieving the resource anyway. You should write code accordingly.
I like urllib2, sample code:
import urllib2
res = urllib2.urlopen('http://google.com/index.html')
res.getCode() #contains code
I something went wrong, you'll get an exception you can catch.
EDIT: Thanks, changes res.code to res.getCode() since the second one is documented
When I run this:
import urllib
feed = urllib.urlopen("http://www.yahoo.com")
print feed
I get this output in the interactive window (PythonWin):
<addinfourl at 48213968 whose fp = <socket._fileobject object at 0x02E14070>>
I'm expecting to get the source of the above URL. I know this has worked on other computers (like the ones at school) but this is on my laptop and I'm not sure what the problem is here. Also, I don't understand this error at all. What does it mean? Addinfourl? fp? Please help.
Try this:
print feed.read()
See Python docs here.
urllib.urlopen actually returns a file-like object so to retrieve the contents you will need to use:
import urllib
feed = urllib.urlopen("http://www.yahoo.com")
print feed.read()
In python 3.0:
import urllib
import urllib.request
fh = urllib.request.urlopen(url)
html = fh.read().decode("iso-8859-1")
fh.close()
print (html)