django/python How to connect to a webservice by posting a json

Question

I am developing an application that connects to a web service, providing a json string with data and receiving a reply. I use the following code, where I build the json and try to post it:

def connectToService(request):
    data='foxp3 factor'
    l=[] 
    l.append(data)
    l.append(80)
    l.append(5)
    data=json.dumps({"findCitations":l})
    result = urllib2.urlopen('http://www.example.com/webservice', urllib.urlencode(data))

But it doesn't work. I hope that the json reply from the web service will be stored in result and then I will figure out a way to parse it, probably by deseriazizing it. Although there is much literature about it (json, simplejson, HttpPequest) and it has to be pretty simple I have not manage to do it yet. Any solutions?

Jon Clements · Accepted Answer · 2013-01-11 11:24:08Z

3

Why not you are using the requests library

Like

payload = {'key1': 'value1', 'key2': 'value2'}
>>> r = requests.post("http://www.example.com/webservice", data=payload)
>>> print r.text

Where payload is the parameter that you are passing .

Hope this will give you an idea

edited Jan 11, 2013 at 11:24

Jon Clements

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answered Jan 11, 2013 at 11:21

masterofdestiny

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3 Comments

geompalik Over a year ago

I think it is very enlightening. Thank you very much. I will try it!

geompalik Over a year ago

@Jon Clements,I tried that but I have not achieved my target yet. I tried this in shell. After performing the request in 'print r.text' it returns a link and no data, which is exactly the same link I see when I put the url in the browser. From this I assume that the service didn't send any data back? I also use headers={"Content-type":"application/json", "encoding":"utf-8"} in the r=requests.post(url, data=json.dumps(payload), headers=headers). Any comment on this? I also try to find another api available to the web to test my code. Do you have something in mind?

geompalik Over a year ago

Ok I did it. I called the service again, but this time with the url that returned in the first call and it worked. Thanks a lot for the advise.

Torkashvand · Accepted Answer · 2016-02-22 18:57:06Z

You can use this code and some error handling idea:

    payload = {'key1': 'value1', 'key2': 'value2'}
    url = "http://www.example.com/webservice"

    try:
        response = requests.post(url, data=payload)

    except requests.exceptions.ConnectionError:
        message = 'This is not the domain we are looking for. URL is: %s' % url
        print e
        sys.exit(1)

    except requests.exceptions.ConnectTimeout:
        message = 'Too slow connection! URL is: %s' % url
        print e
        sys.exit(1)

    except Exception as e:
        message = 'Unknown Error: %(message)s URL is: %(url)s' % {'message': str(e), 'url': url}
        print e
        sys.exit(1)

    else:
        return response.json()

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