I'm using the Requests: HTTP for Humans library and I got this error:
No connection adapters were found for '192.168.1.61:8080/api/call'
What does this mean and how can I fix it?
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8 Answers
You need to include the protocol scheme:
'http://192.168.1.61:8080/api/call'
Without the http:// part, requests has no idea how to connect to the remote server.
Note that the protocol scheme must be all lowercase; if your URL starts with HTTP:// for example, it won’t find the http:// connection adapter either.
answered Feb 27, 2013 at 15:01
Martijn Pieters
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Comments
Maybe your URL includes some hidden characters, such as \n.
If you define your URL like below, the exception you have seen will be raised:
url = '''
http://google.com
'''
because there are \n hidden in the string. The URL in fact becomes:
\nhttp://google.com\n
4 Comments
Christian Long
Or if your url is accidentally a tuple because of a trailing comma
url = self.base_url % endpoint,
Ravi Shankar Bharti
@ChristianLong is there any way to convert a string to proper url? Like, can you tell me, what are you doing in your comment?
Intelligent-Infrastructure
for me it was the ` " "` not sure why, but at some point my single-quoted URL became single-quoted and then double-quoted (maybe pyCharm, maybe github, at some point it must got filtered that way. I had "google.com" and then it became " ' google.com ' " - after making it just a string it worked again.
Little Bobby Tables
A trailing comma made my url a tuple... Thanks for getting to me check the silly stuff!
I received this error when I refactored a URL, leaving an erroneous comma, thus converting my URL from a string into a tuple.
My exact error message:
741 # Nothing matches :-/
--> 742 raise InvalidSchema("No connection adapters were found for {!r}".format(url))
743
744 def close(self):
InvalidSchema: No connection adapters were found for "('https://api.foo.com/data',)"
Here's how that error came to be born:
# Original code:
response = requests.get("api.%s.com/data" % "foo", headers=headers)
# --------------
# Modified code (with bug!)
api_name = "foo"
url = f"api.{api_name}.com/data", # !!! Extra comma doesn't belong here!
response = requests.get(url, headers=headers)
# --------------
# Solution: Remove erroneous comma!
api_name = "foo"
url = f"api.{api_name}.com/data" # No extra comma!
response = requests.get(url, headers=headers)
Comments
As stated in a comment by christian-long
Your url may accidentally be a tuple because of a trailing comma
url = self.base_url % endpoint,
Make sure it is a string
3 Comments
Pfalbaum
changing the tuple to a string gave me the same error. Can you give an example?
Gulzar
Not sure what you mean by changing to string. I meant
url = self.base_url % endpoint without the comma
Dr.CKYHC
Similar case happens by me: I have incidentally placed scopes around the url, e.g.
my_url = ['https://my_server/my_link/' + 'rest/api/2/search/...'] In the error-message it's quite hard to recognize - what is the root cause (you can consider scopes as value formatting). But the one with comma is more tricky ;-)In my case the issue was that I had defined URLs as a value to an environment variable in an .env, like this:
MY_ENDPOINT="<url>"
The problem with this is that, at least Docker, takes the quotes as part of the URL string, making the URL invalid. Removing the quotes solved the issue for me:
MY_ENDPOINT=<url>
Comments
I was trying to call json-server with requests, however I was unable to call the API.
The root problem was, it requires to be included the protocol scheme i.e. http://
import requests
url = 'http://localhost:3000/employees'
try:
response = requests.get(url)
if response:
print(response.json())
except requests.exceptions.RequestException as ex:
print(ex)
Also, for http requests - the requests should be wrapped with try/except block - to better debug the cause of the problem.
Comments
For me the issue was I was passing url as below on windows:
MY_ENDPOINT="<url>"
I had to change that to below without greater than and less than symbol to get rid of this error.
MY_ENDPOINT="url"
Comments
In my case I used default url for method
def call_url(url: str = "https://www.google.com"):
and url attr was overridden by some method to /some/url
