我想连接一个动作,如果手势是一个轻击,它会以特定方式动画一个对象,但如果按下持续时间超过.5秒,它会做其他事情.
现在,我只是把动画连接起来了.我不知道如何区分长按和水龙头?如何访问印刷机持续时间以实现上述目标?
@IBAction func tapOrHold(sender: AnyObject) {
UIView.animateKeyframesWithDuration(duration, delay: delay, options: options, animations: {
UIView.addKeyframeWithRelativeStartTime(0, relativeDuration: 0, animations: {
self.polyRotate.transform = CGAffineTransformMakeRotation(1/3 * CGFloat(M_PI * 2))
})
UIView.addKeyframeWithRelativeStartTime(0, relativeDuration: 0, animations: {
self.polyRotate.transform = CGAffineTransformMakeRotation(2/3 * CGFloat(M_PI * 2))
})
UIView.addKeyframeWithRelativeStartTime(0, relativeDuration: 0, animations: {
self.polyRotate.transform = CGAffineTransformMakeRotation(3/3 * CGFloat(M_PI * 2))
})
}, completion: { (Bool) in
let vc : AnyObject! = self.storyboard?.instantiateViewControllerWithIdentifier("NextView")
self.showViewController(vc as UIViewController, sender: vc)
})
Run Code Online (Sandbox Code Playgroud) 我目前正在这样做:
if x in a and y in a and z in a and q in a and r in a and s in a:
print b
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是否有更多的pythonic方式来表达这种if说法?
我在ECS上有一个回购,使用ecs-cli创建了一个集群
ecs-cli configure --region=us-west-2 --profile=<MY PROFILE> --cluster=cluster-1
ecs-cli up --capability-iam --keypair=<MY KEYPAIR>
但是,执行compose文件的下一步是失败时
ecs-cli compose --file docker-compose.yml --project-name drafter-project service up
这是我的docker-compose.yml文件:
version: '2'
services:
rabbit:
image: rabbitmq
hostname: rabbit1
ports:
- 5672:5672
- 15672:15672
drafter:
build: .
depends_on:
- rabbit
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我得到的错误是:
Error registering task definition
error=ClientException: Container.image should not be null or empty.
Create task definition failed
error=ClientException: Container.image should not be null or empty.
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我不确定什么是任务定义或它需要什么.
我不能让我的芹菜工人经常听默认队列.芹菜不断退出.
$: docker-compose up
Starting tasker_rabbitmq_1
Starting tasker_celery_1
Attaching to tasker_rabbitmq_1, tasker_celery_1
tasker_celery_1 exited with code 1
rabbitmq_1 |
rabbitmq_1 | RabbitMQ 3.6.1. Copyright (C) 2007-2016 Pivotal Software, Inc.
rabbitmq_1 | ## ## Licensed under the MPL. See http://www.rabbitmq.com/
rabbitmq_1 | ## ##
rabbitmq_1 | ########## Logs: /var/log/rabbitmq/rabbit@0bcd2c4762eb.log
rabbitmq_1 | ###### ## /var/log/rabbitmq/rabbit@0bcd2c4762eb-sasl.log
rabbitmq_1 | ##########
rabbitmq_1 | Starting broker... completed with 6 plugins.
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我正在尝试构建一个具有单独作业层的应用程序作为单独部署的作业容器.所以架构是:
这就是我所拥有的:
文件夹结构:
-tasker
-tasker …Run Code Online (Sandbox Code Playgroud) 如何在docker容器中运行Celery和RabbitMQ?你能指点我试试dockerfile或撰写文件吗?
这就是我所拥有的:
Dockerfile:
FROM python:3.4
ENV PYTHONBUFFERED 1
WORKDIR /tasker
ADD requirements.txt /tasker/
RUN pip install -r requirements.txt
ADD . /tasker/
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泊坞窗,compose.yml
rabbitmq:
image: tutum/rabbitmq
environment:
- RABBITMQ_PASS=mypass
ports:
- "5672:5672"
- "15672:15672"
celery:
build: .
command: celery worker --app=tasker.tasks
volumes:
- .:/tasker
links:
- rabbitmq:rabbit
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我遇到的问题是我不能让芹菜活着或跑步.它不断退出.
我的S3存储桶是私有的。但是现在我已将其更新为公开。我似乎仍然无法在未签名URL等的情况下访问文件。如何将所有文件公开?
我正在使用以下代码来发帖。
var checked = ["2231","2432"];
jQuery.ajax({
type: 'post',
url: statusUrl,
data: {"entries":checked},
...
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然而,当它实际发布时,服务器端的发布数据和开发人员工具中的检查始终是
{entries[]: 2342 etc}
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为什么会出现方括号?我怎样才能摆脱它?
我的模型有一个CharField默认为随机生成的字符串,该字符串应该是唯一的。我不想使用UUIDField. IntegrityError如果生成的代码是重复的,有没有办法覆盖 save 方法来处理引发的问题?或者我应该在我的生成函数中检查它吗?
模型
class Item(models.Model):
...
item_code = models.CharField(max_length=11, default=get_generated_code, unique=True)
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功能
def get_generated_code():
code = ''.join(random.SystemRandom().choice(string.ascii_letters + string.digits) for _ in range(11))
return code
Run Code Online (Sandbox Code Playgroud) 我在我的app中使用了一个custon用户模型fowl.当我运行syncdb或者makemigrations还是migrate我收到了LookupError.请帮忙
在settings.py我定义AUTH_USER_MODEL为'fowl.User'
禽/ models.py
from django.db import models
from django.contrib.auth.models import AbstractBaseUser, PermissionsMixin, BaseUserManager
from django.utils import timezone
from django.core.mail import send_mail
from django.utils.translation import ugettext_lazy as _
class UserManager(BaseUserManager):
def create_user(self, email, password=None):
"""
Creates and saves a User with the given email, date of
birth and password.
"""
if not email:
raise ValueError('Users must have an email address')
user = self.model(
email=self.normalize_email(email),
) …Run Code Online (Sandbox Code Playgroud) python django django-models django-authentication django-1.8
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