【问题标题】:Getting 200 ok status but giving response error from postman获得 200 ok 状态但从邮递员那里得到响应错误
【发布时间】:2018-11-05 21:15:22
【问题描述】:

得到响应 200 ok 状态但从邮递员那里得到响应错误 霸气的

 {
        "errors": "Unable to log you in, please try again.",
        "success": false
    }

发帖网址:https://demo.cognitonetworks.com/cognito/gettoken

[{"key":"Content-Type","value":"application/json","description":""}]

@csrf_exempt
    def token_new(request):      
    if request.method == 'POST':
        email = request.POST.get('username')
        print email
        try:
            UserObj = CognitoUser.objects.get(user__email=email)
            username =  UserObj.user.username
            group = UserObj.user.groups.filter(name__in=['Admin','Manager'])
            group_name = ''
            if group:
                group_name = group[0].name                                                     
        except:
            return JsonResponseUnauthorized("Unable to log you in, please try again.")                    
        password = request.POST.get('password')

        if username and password:
            user = authenticate(username=username, password=password)
            if user:
                TOKEN_CHECK_ACTIVE_USER = getattr(settings, "TOKEN_CHECK_ACTIVE_USER", False)
                if TOKEN_CHECK_ACTIVE_USER and not user.is_active:
                    return JsonResponseForbidden("User account is disabled.")
                token = token_generator.make_token(user)
                data = {
                    'token': token,
                    'user': user.pk,
                    'userName':UserObj.user.username,
                    'companyId':UserObj.company.companyid,
                    'companyApikey':UserObj.company.apikey,
                    'group_name':group_name
                }
                request.session['token'] = token
                request.session['token'] = user.pk
                return JsonResponse(data)
            else:
                return JsonResponseUnauthorized("Unable to log you in, please try again.")
        else:
            return JsonError("Must include 'username' and 'password' as POST parameters.")
    else:
        return JsonError("Must access via a POST request.")

【问题讨论】:

    标签: javascript python html ajax django


    【解决方案1】:

    答案很简单,服务器只是接受你的请求,然后给你一个json

    { "errors": "Unable to log you in, please try again.", "success": false }
    

    问题不在客户端或服务器上,您应该阅读:

    无法登录,请重试

    【讨论】:

    • 可能是什么问题
    • 身份验证错误
    • 但是我可以使用给定的用户名和密码登录...我在这里通过相同的身份验证?
    猜你喜欢
    • 2020-02-09
    • 2014-08-03
    • 1970-01-01
    • 2020-11-25
    • 1970-01-01
    • 1970-01-01
    • 2019-04-01
    • 2020-05-05
    • 1970-01-01
    相关资源
    最近更新 更多