小编Sus*_*S N的帖子

WARNING Application instance <Task pending coro=<__call__() running at 
/home/developer/projects/tabcon/tabcon_env/lib/python3.5/site-packages/channels/http.py:191>
wait_for=<Future pending cb=[_chain_future.<locals>._call_check_cancel() at /usr/lib/python3.5/asyncio/futures.py:431, Task._wakeup()]>> 
for connection <WebRequest at 0x7ffa40e17b00 method=GET uri=/ clientproto=HTTP/1.1> took too long to shut down and was killed. 

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出现上述异常，整个服务停止运行。

我在用：

• 频道==2.0
• channel_redis==2.2.1
• 达芙妮==2.2.2
• 姜戈==2.1
• channel_redis==2.2.1

任何想法,为什么以下脚本的下拉菜单不会保留在选定的值上:

 if (isset($_POST['Submit1'])) {
     $frequency = $_POST['frequency'];
     $fileName = 'ConfigurationFile.txt';

     $lines = file($fileName);

     $lines[0]="Center_frequency= ". $frequency."\n";

     file_put_contents($fileName, implode($lines));

 } else {

     $fileName = 'ConfigurationFile.txt';

     $lines = file($fileName);

     $frequency=explode(" ",$lines[0]);
     $frequency=$frequency[1];
 }
 ?>

但如果我把我的号码放在一个引号中

 <?php

 if (isset($_POST['Submit1'])) {
     $frequency = $_POST['frequency'];
     $fileName = 'ConfigurationFile.txt';

     $lines = file($fileName);

     $lines[0]="Center_frequency= ".'"'. $frequency.'"'."\n";

     file_put_contents($fileName, implode($lines));

 } else {

     $fileName = 'ConfigurationFile.txt';

     $lines = file($fileName);

     $frequency=explode("\"",$lines[0]);
     $frequency=$frequency[1];
 }
 ?>

它会像魅力一样工作,意味着选定的值将保留,下拉菜单不会转到列表中的第一个值.这是下拉菜单的html代码

     <form action="test.php" name="Calculation" method="post">
     Center Frequency: 
     <select name="frequency" >

     <?php $attr= 'selected="selected"'; ?>

    <option value="4.463" …

视图.py

class CountryViewSet(viewsets.ViewSet):   
    serializer_class = CountrySerializer
    pagination_class = LimitOffsetPagination
    def list(self,request):
        try:
            country_data = Country.objects.all()
            country_serializer = CountrySerializer(country_data,many=True)
            return Response(            
                data = country_serializer.data,
                content_type='application/json',            
                )
        except Exception as ex:
            return Response(
                data={'error': str(ex)},
                content_type='application/json',
                status=status.HTTP_400_BAD_REQUEST
                )

我添加的 Settings.py

'DEFAULT_PAGINATION_CLASS': 'rest_framework.pagination.LimitOffsetPagination',

在我的 urls.py

router = routers.DefaultRouter(trailing_slash=False)

router.register(r'country', CountryViewSet, base_name='country')
urlpatterns = [
    url(r'^', include(router.urls)),
]

当我尝试使用此 URL http://192.168.2.66:8001/v1/voucher/country 时，它正在返回所有数据。

但是当我尝试使用这个 URL http://192.168.2.66:8001/v1/voucher/country/?limit=2&offset=2

但它返回 404 错误。我是 django 的新手。请帮助我:)

我一直在尝试 youtube content api 并尝试使用以下代码

def main():
  
  credentials = ServiceAccountCredentials.from_json_keyfile_name(
    filename='bbtv-306609-b20e12d67493.json',
    scopes=['https://www.googleapis.com/auth/youtubepartner',
    'https://www.googleapis.com/auth/youtube.force-ssl',
    'https://www.googleapis.com/auth/youtubepartner'])
  http = httplib2.Http()
  http = credentials.authorize(http)
  service = build("youtubePartner", version="v1", http=http)

最终以..

我在成员下添加了以下用户