我会将多个任务添加到芹菜队列并等待结果.我有各种想法如何利用某种形式的共享存储(memcached,redis,db等)实现这一点,但是,我会认为这是Celery可以自动处理但我无法在线找到任何资源的东西.
代码示例
def do_tasks(b):
for a in b:
c.delay(a)
return c.all_results_some_how()
我已经编写了以下代码来使用boto对URL进行签名,但是,我无法访问已签名的URL.任何帮助将不胜感激.
con = cloudfront.CloudFrontConnection(
settings.AWS_USER_DETAILS.access_key_id,
settings.AWS_USER_DETAILS.secret_key_id,
)
identities = con.get_all_origin_access_identity()
if len(identities) < 1:
oai = con.create_origin_access_identity(
'cv-oai',
'OAI signing iOS video requests'
)
else:
oai = con.get_all_origin_access_identity()[0]
distro_summary = con.get_all_distributions()[0]
distro_info = con.get_distribution_info(distro_summary.id)
bucket = boto.connect_s3(
settings.AWS_USER_DETAILS.access_key_id,
settings.AWS_USER_DETAILS.secret_key_id,
host='s3-eu-west-1.amazonaws.com'
).get_bucket(u'cv-video')
key_str = video.video_url.split('/')[-1]
key = bucket.get_key(key_str)
key.add_user_grant("READ", oai.s3_user_id)
distro = distro_summary.get_distribution()
# For local testing
if ip_addr == '127.0.0.1':
ip_addr = None
private_key_string = open(settings.CF_PRIV_KEY_LOC).read()
print private_key_string
signed_url = distro.create_signed_url(
"https://%s/%s" % (distro_info.domain_name, key_str),
'APKAI3I3QZ7I73OIQIXA',
expire_time=int(time.time()) + video.video_length + …我正在使用django-tastypie编写 API 。我有两个自定义权限问题，希望django-guardian能够解决。
我有两个用户组临床医生和患者。临床医生应该能够访问仅属于其患者的对象，并且患者应该只能访问他们自己创建的对象。
我的代码如下：
class UserResource(ModelResource):
class Meta:
queryset = User.objects.all()
resource_name = 'auth/user'
excludes = ['email', 'password', 'is_superuser']
class BlogPostResource(ModelResource):
author = fields.ToOneField(UserResource, 'author', full=True)
class Meta:
queryset = BlogPost.objects.all()
resource_name = 'posts'
allowed_methods = ["get", "post"]
# Add it here.
authentication = BasicAuthentication()
authorization = DjangoAuthorization()
filtering = {
'author': ALL_WITH_RELATIONS,
}
我如何使用权限来限制对此的访问BlogPostResource？
AAAAAAFuAAIAAAZNYWMgT1MAAAAAAAAAAAAAAAAAAAAAAAAAAADMrsHTSCsAAAALuG8NYWxleHN1Y2tzLmRpYwAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAPJXS83SjIoAAAAAAAAAAP////8AAAkgAAAAAAAAAAAAAAAAAAAAB0Rlc2t0b3AAABAACAAAzK6zwwAAABEACAAAzdJ+egAAAAEADAALuG8AC7hIAADK3wACADFNYWMgT1M6VXNlcnM6AGFuZHJld3ByeWRlOgBEZXNrdG9wOgBhbGV4c3Vja3MuZGljAAAOABwADQBhAGwAZQB4AHMAdQBjAGsAcwAuAGQAaQBjAA8ADgAGAE0AYQBjACAATwBTABIAJ1VzZXJzL2FuZHJld3ByeWRlL0Rlc2t0b3AvYWxleHN1Y2tzLmRpYwAAEwABLwAAFQACABL//wAA
它是~/Library/Preferences/com.microsoft.office.plistMicrosoft Office 2011 Mac文件中的数据字段.
它使用base64进行部分解码,但似乎不完全是base64.
编辑: 这是另一个例子.
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
<plist version="1.0">
<data>AAAAAAFWAAIAAAZNYWMgT1MAAAAAAAAAAAAAAAAAAAAAAAAAAADMrsHTSCsAAAALuG8HMm5kLmRpYwAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAPgev83SjIoAAAAAAAAAAP////8AAAkgAAAAAAAAAAAAAAAAAAAAB0Rlc2t0b3AAABAACAAAzK6zwwAAABEACAAAzdJ+egAAAAEADAALuG8AC7hIAADK3wACACtNYWMgT1M6VXNlcnM6AGFuZHJld3ByeWRlOgBEZXNrdG9wOgAybmQuZGljAAAOABAABwAyAG4AZAAuAGQAaQBjAA8ADgAGAE0AYQBjACAATwBTABIAIVVzZXJzL2FuZHJld3ByeWRlL0Rlc2t0b3AvMm5kLmRpYwAAEwABLwAAFQACABL//wAA</data>
</plist>
披露:这是大学工作.我不希望代码完成我的任务,而是想了解如何最好地操作C中的字符串.
我需要编写一个函数来以某种方式操作字符串(我不是说什么,以确保没有人提供确切的代码).
在python中,我只是做以下事情
def foo(str):
return str
显然事情并不像C中那么容易.
谁能告诉我如何最好地实现这一目标.我应该使用指针来模拟通过引用传递并只是操纵原始字符串吗？任何帮助/资源将不胜感激.
更新:我确实想对字符串执行一个操作并返回该操作的结果(也是一个字符串).我很乐意操纵原始字符串或返回它.哪个曾被认为是最佳实践.
我设置的任务是基于如何进行操作,所以我不想明确这样做.
所以Python将是:
def foo(str):
#do something to str (which doesn't change it's length)
return str