小编Sye*_*b M的帖子

正如我在这里提到的,我创建了View.

是否可以创建用于会话的视图类？

v = Table('viewname', metadata, autoload=True)

class ViewName(object):
    def __init__(self, name):
       self.name = name

mapper(ViewName, v)

我只想在给出query_params的情况下返回model_info,否则它应该给出一些错误消息.

我试过下面的代码,但是当查询参数没有传递名称时,它给了我keyerror.

from rest_framework.validators import ValidationError

class ModelSerializer(serializers.ModelSerializer):
    class Meta:
        model = ModelName
        fields = ('name', 'abbreviation')

    def validate_name(self, value):
        if value:
            return value
        else:
            raise ValidationError('Enter name')



class ModelNameListList(generics.ListCreateAPIView):
    renderer_classes = (JSONRenderer, )
    serializer_class = ModelSerializer

    def get_queryset(self):
        queryset = ModelName.objects.all()
        name = self.request.query_params['name']
        queryset = queryset.filter(Q(name__icontains=name) | Q(abbreviation__icontains=name)).all()
        return queryset

我不能使用get方法,因为我也使用分页,如果我将使用get方法,它将返回所有结果.

当我在get_queryset中使用下面的代码时,我得到的响应对象没有长度

   def get_queryset(self):
        queryset = ModelName.objects.all()
        name = self.request.query_params.get('name', None)
        if name:
            queryset = queryset.filter(Q(name__icontains=name) | Q(abbreviation__icontains=name)).all()
        else:
            content = {'errors': 'name is missing'}
            return Response(content)
        return …

我正在尝试使用python-boto配置的多个网络在opensatck设置上启动实例.

但我得到了以下错误,

EC2ResponseError: EC2ResponseError: 400 Bad Request <?xml version="1.0"?> <Response><Errors><Error><Code>NetworkAmbiguous</Code><Message>Multiple possible networks found, use a Network ID to be more specific.</Message></Error></Errors><RequestID>req-28b5a4e8-3838-4111-95db-337c5048716d</RequestID></Response>

我的代码就像这里,

from boto import ec2
ostack = ec2.connection.EC2Connection(
    ec2_access_key, ec2_secret_key, is_secure=False, port=8773, region='nova',
    path='/services/Cloud'
)

ostack.run_instances('ami-xxxxx', key_name='BotoTest')

以上适用于配置为openstack的单个网络.

注意:run_instances没有network-id的关键字参数.

我犯了什么错误或如何解决它？或者它是python-boto中的错误？

谢谢你.

我在django app views.py中有这样的课程

from dms.models import Folder, File, FileTag

class GetFamily:

    def getFathers(folder_id):
        if folder_id == None:
            rev_fathers=None
        else:
            fathers=[]
            rev_fathers=[]
            father=Folder.objects.get(id=folder_id)
            fathers.append(father)
            while father.parent_folder_id != None:
                father=Folder.objects.get(id=father.parent_folder_id)
                fathers.append(father)
            rev_fathers=reversed(fathers)
        return rev_fathers

    def getChildrenFolders(folder_id):
        folders=Folder.objects.filter(parent_folder_id=folder_id)
        return folders  

    def getChildrenFiles(folder_id):
        files=File.objects.filter(folder_id=folder_id)
        return files

当我用id调用getFathers(folder_id)方法时,它给getFathers()取1个参数(给定2个)

请帮助我

我有以下代码片段async/await.

async function test1 () {
  setTimeout(() => {
    console.log("1")
  }, 2000);
}

async function test2 () {
  setTimeout(() => {
    console.log("2")
  }, 1000);
}

async function test3 () {
  setTimeout(() => {
    console.log("3")
  }, 1500);
}

async function test4 () {console.log("4")}

async function run () {
  await test1()
  await test2()
  await test3()
  await test4()
}

run()

当我探索上面的代码片段时,我期待输出为1,2,3,4.但我得到了4,2,3,1.我错过了什么吗？

节点版本 v10.13.0

我想在txt文件中保存函数调用的值,但我得到的是typeerror:期望一个字符缓冲区对象

def printme( x ):
    "Function f(x) = x(x+2)"
    print x*(x+2)
    return

myfile = open('xyz.txt', 'w')
    for w in b:
        myfile.write('%d' (printme(w)))

myfile.close()