Python GET is not working in browser

Question

I have a Django custom URL which works fine in POSTMAN but not working properly in a browser the details are given as below.

In postman, I am using the following URL and it s working fine 127.0.0.1:8000/v0/call_letter_status/ and I am getting a 200 response and output as well

But when I am trying in browser I am getting an error like this

ValueError at /v0/call_letter_status/

The view project.views.User.call_letter_track didn't return an HttpResponse object.

    Request Method: GET

    Request URL:    http://127.0.0.1:8000/v0/call_letter_status/

    Django Version: 1.5

    Exception Type: ValueError

My Code is as given below:

def call_letter_track(request):
     if request.META["CONTENT_TYPE"] == 'application/json':
         if request.method == 'GET':
             sqlQuery = """ SELECT jc.company_name,jc.job_position,jc.venue,jc.email_body,jc.interview_date,aj.job_id,aj.logo_image_url FROM jr_call_letter jc
                        JOIN api_job aj ON aj.job_id=jc.job_id ORDER BY "jc.job_id" DESC LIMIT 2 """

             cursor.execute(sqlQuery)
             result=dictfetchall(cursor)
             final_response_map = []
             key=0
             for result_new in result:
                print key
                response_map = {}
                response_map['company_name']=result[key]['company_name'] 
                response_map['job_id']=result[key]['job_id']  
                response_map['job_position']=result[key]['job_position'] 
                response_map['interview_date']=datetime.fromtimestamp(result[key]['interview_date']).strftime('%d-%m-%Y')
                response_map['email_body']=result[key]['email_body']
                response_map['venue']=result[key]['venue']
                response_map['logo_image_url']=result[key]['logo_image_url']
                key=key+1
                final_response_map.append(response_map)
             response = {'data':final_response_map}
             data = json.dumps(response, encoding="ISO-8859-1")  
         return HttpResponse(data,content_type="application/json", status=200)

Please help me in getting a solution for this problem.

Arpit Goyal · Accepted Answer · 2015-07-31 09:18:45Z

2

Your return statement is inside the if condition. If that condition is invalid it will go outside of the condition and expect a Response, but there is no return outside your condition, hence the error.

try providing this for checking:

def call_letter_track(request):
    if request.META["CONTENT_TYPE"] == 'application/json':
    '''
    .
    .
    your code
    .
    .
    '''
        return HttpResponse(data,content_type="application/json", status=200)
    return HttpResponse('Hello World')

The browser by default has the Content-Type header of application/xml and hence it is not entering your if condition.

answered Jul 31, 2015 at 9:18

Arpit Goyal

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falsetru · Accepted Answer · 2015-07-31 09:19:18Z

1

Browser will not send CONTENT_TYPE header with application/json; causes the outer if block is never executed; The view function will not return.

How about Remove the outermost if so that request without Content-type: application/json also get HttpResponse?

def call_letter_track(request):
     if request.META["CONTENT_TYPE"] == 'application/json':  # <---
         if request.method == 'GET':
             ....

answered Jul 31, 2015 at 9:19

falsetru

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2 Comments

thomas chacko Over a year ago

#falsetru This one actually works but the problem with this is that,If I want to pass an ID and want to get the result for that ID as o/p then I should need to use this right???

falsetru Over a year ago

@thomaschacko, Do you mean passing as a query string like ....?id=something ? Then, you can access the id using request.REQUEST.get('id') or request.GET.get('id')

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