I am new to python programming. I am trying to access LinkedIn data using pyhton.
I have performed all the steps mentioned at https://ozgur.github.io/python-linkedin/
When I run the code to get profile, i get the desired result but if I try to code anything else (like search_profile() or get_group(41001)) I get the following error
LinkedInError: 400 Client Error: Bad Request for url:
The code that have written is below:
API_KEY = "xyz"
API_SECRET = "abc"
RETURN_URL = "http://localhost:8080/testPrj/auth/linkedin"
Access_Token = "abcdef"
application = linkedin.LinkedInApplication(token=Access_Token)
application.get_profile() # No Error
application.get_group(41001) # Error
application.search_profile(selectors=[{'people':['first_name','last_name']}]) # Error
Can someone help me to understand what is this issue and how to solve it ?
Thanks !!!
Related
I've been working on a django project that needs to send faxes.
For sending faxes I am using interfax python library.
To generate pdf from html, I am using xhtml2pdf.
I wrote like below, and it didn't work and threw an error.
I don't know what to do now.
Please help.
The code
# interfax authentication
interfax_password = config("INTERFAX_PASSWORD")
interfax_account = config("INTERFAX_ACCOUNT")
interfax = InterFAX(username=interfax_account, password=interfax_password)
f = File(interfax, pdf, mime_type="application/pdf")
fax_number = config("INTERFAX_DESTINATION")
# actually sending the data
fax = interfax.outbound.deliver(fax_number=fax_number, files=[f])
The error thrown
equests.exceptions.HTTPError: 400 Client Error: Bad Request for url:https://rest.interfax.net/outbound/faxes?faxNumber=111111111
Thank you in advance
I think I just need to use the API directly.
I see there has been questions similar to this but the responses to them explains my asking a similar question here. I am wanting to utilize LinkedIn's REST API for my personal job hunt, specifically utilizing the job search API capabilities.
My problem comes in with retrieving an access token and autorisation to actually use the app. I have tried the below code - the first cell is to retrieve autorisation and the second is to retrieve an access token:
from linkedin import linkedin
APPLICATON_KEY = 'XXXXXX'
APPLICATON_SECRET = 'XXXXX'
RETURN_URL = 'http://localhost:8000'
authentication = linkedin.LinkedInAuthentication(APPLICATON_KEY, APPLICATON_SECRET, RETURN_URL,
linkedin.PERMISSIONS.enums.values())
print (authentication.authorization_url) #open this url on your browser
Access token:
authentication = linkedin.LinkedInAuthentication(
APPLICATON_KEY,
APPLICATON_SECRET,
RETURN_URL,
linkedin.PERMISSIONS.enums.values()
)
authentication.authorization_code = '#############################################'
result = authentication.get_access_token()
print ("Access Token:", result.access_token)
print ("Expires in (seconds):", result.expires_in)
When I attempt to retreieve authorisation (required for access token) the following error occurs:
File "/usr/local/lib/python3.6/dist-packages/linkedin/linkedin.py", line 294
except (requests.ConnectionError, requests.HTTPError), error:
^
SyntaxError: invalid syntax
It's an obvious syntax error, however am ignorant to what it is exactly - I have a feeling it might be my localhost
That error points to the comma separating the exception classes from the variable name.
A comma there is Python 2.x syntax, so that means the linkedin library you're using is not compatible with Python 3.
If you're using this library, you can see it's not been updated in 5 years. This (PyPI) seems like a slightly fresher fork.
I have this function in Python to get media id of the post from its URL provided:
def get_media_id(self):
req = requests.get('https://api.instagram.com/oembed/?url={}'.format(self.txtUrl.text()))
media_id = req.json()['media_id']
return media_id
When I open the result URL in the browser it returns data but in the code the result is "404 not found"
For example consider this link:
https://www.instagram.com/p/B05bitzp15CE8e3Idcl4DAb8fjsfxOsSUYvkDY0/
When I put it in the url the result is:
But when I run the same in this function it returns 404 error
I tried running your code and assuming that self.txtUrl.text() there is nothing wrong with your code. The problem is that you are trying to get access to a media id of a private account without the access token.
The reason that you are able to open that link in the browser is due to the fact that you are likely logged into that account or are following it. To use the method you have given, you would need a public instagram post, for example try setting txtUrl.text() = https://www.instagram.com/p/fA9uwTtkSN/. Your code should work just fine.
The problem is that your GET request doesn't have any authorisation token to gain access to the post. Other people have written answers to how to get the media_id if you have the access token here: Where do I find the Instagram media ID of a image (having access token and following the image owner)?
I'm writing an application that utilizes Paypal's permissions API. I'm currently working on the sandbox. I get the verification code correctly but when I try to GetAccessToken, I get the error:
{"responseEnvelope":{"timestamp":"2013-09-03T08:32:16.580-07:00","ack":"Failure","correlationId":"3527b7033f20f","build":"2210301"},"error":[{"errorId":"560022","domain":"PLATFORM","subdomain":"Application","severity":"Error","category":"Application","message":"The X-PAYPAL-APPLICATION-ID header contains an invalid value","parameter":["X-PAYPAL-APPLICATION-ID"]}]}
I'm using the sandbox APP_ID and all the Verification code is also gotten dynamically. Here is my code fragment.
token = "AAAAAAAYaraTSVjvkUBT"
verification = "mgnnWDVfFmgAES0q371Hug"
headers2 = {
"X-PAYPAL-SECURITY-USERID": settings.USERNAME,
"X-PAYPAL-SECURITY-PASSWORD": settings.PASSWORD,
"X-PAYPAL-SECURITY-SIGNATURE": settings.SIGNATURE,
"X-PAYPAL-REQUEST-DATA-FORMAT": "JSON",
"X-PAYPAL-RESPONSE-DATA-FORMAT": "JSON",
"X-PAYPAL-APPLICATION-ID": "APP-80W284485P519543T",
}
url = "https://svcs.paypal.com/Permissions/GetAccessToken/?token=%s&verifier=%s" %(token, verification)
dat2 = {"requestEnvelope": {"errorLanguage":"en_US"}}
req2 = urllib2.Request(url, simplejson.dumps(dat2), headers2)
res2 = urllib2.urlopen(req2).read()
What I'm I doing wrong??
You cannot use the sandbox application id on the live environment. See https://developer.paypal.com/webapps/developer/docs/classic/lifecycle/goingLive/#register to learn how to obtain a live application id.
The endpoint should be https://svcs.sandbox.paypal.com as Siddick said above. The paypal API documentation is so inconsistent, the endpoint i had used previously had been used in a sandbox situation in the documentation.
I'm making a web app in python using the Flask framework to request the access token from Facebook using the SDK supplied in their site.
The access token is returned and it is correctly set in the GraphAPI object. However, it is returning the following error:
GraphAPIError: Invalid OAuth access token.
If I query the graph API from my local python environment using the same access token, it works just fine. The problem seems to be when executing in the webserver.
See code snippet below:
#app.route('/facebook')
def fb():
if 'token' in session:
graph = facebook.GraphAPI(session['token'])
return graph.get_object("me")
#app.route('/facebook/login')
def fblogin():
code = request.args.get('code','')
if(code == ""):
args = dict(client_id=app_id, redirect_uri=request.base_url)
#args = dict(client_id=app_id, redirect_uri=request.url_root + 'facebook')
return redirect(
"https://graph.facebook.com/oauth/authorize?" +
urllib.urlencode(args))
else:
token = facebook.get_access_token_from_code(code, request.base_url, app_id, app_secret)
session['token'] = [token.itervalues().next()]
return redirect (request.url_root + 'facebook')
Has anyone faced this before and/or can provide some insights?
Ok, 2 issues that I have managed to correct in this code and get it working:
1) The following line of code makes a list, that why the GraphAPI object is not able to identify a valid access token:
session['token'] = [token.itervalues().next()]
2) The following line of code gives an error stating that 'dict' is not callable. This is because the returned variable is a dictionary and, in order to be returned as a view, one must first transform it into a string:
return graph.get_object("me")