我正在尝试从其状态创建用户的活动流.
楷模:
class Status(models.Model):
body = models.TextField(max_length=200)
image = models.ImageField(blank=True, null=True, upload_to=get_upload_file_name)
privacy = models.CharField(max_length=1,choices=PRIVACY, default='F')
pub_date = models.DateTimeField(auto_now_add=True, auto_now=False)
user = models.ForeignKey(User)
class Activity(models.Model):
actor = models.ForeignKey(User)
action = models.CharField(max_length=100)
content_type = models.ForeignKey(ContentType)
object_id = models.PositiveIntegerField()
content_object = generic.GenericForeignKey('content_type', 'object_id')
pub_date = models.DateTimeField(auto_now_add=True, auto_now=False)
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但是,虽然我创建了一个新状态,但它不会根据post_save信号创建新活动.
信号:
from django.contrib.contenttypes.models import ContentType
from django.db.models.signals import post_save
from status.models import Status
from models import Activity
def create_activity_item(sender, instance, signal, *args, **kwargs):
if kwargs.get('created', True):
ctype = ContentType.objects.get_for_model(instance)
if ctype.name == …Run Code Online (Sandbox Code Playgroud) 我正在使用来自react-native 的新React-navigation.我的导航如下:
StackNavigator:
完整代码:
const MainNavigation = StackNavigator({
Home: {
screen: HomeNavigation,
},
Notification: {
screen: NotificationNavigation,
}
});
const HomeNavigation = TabNavigator({
AllPost: {
screen: All,
},
ArticlePost: {
screen: Article,
},
BusinessPost: {
screen: Job,
},
});
HomeNavigation.navigationOptions = {
title: 'Home',
header: {
right: <SearchNotification/>
},
};
class SearchNotification extends React.Component {
goToNotification = () => {
this.props.navigate('Notification');
};
render() {
return (
<View style={styles.container}>
<TouchableOpacity>
<Icon name="md-search" style={styles.Icon}/> …Run Code Online (Sandbox Code Playgroud) 我应该如何创建一个自定义页面过渡,听取轻击和滑动/拖动?.如果页面被向右/向左拖动(同时仍然保持),则将页面拖动/滑动到滑动/阈值.释放水龙头时,如果值等于或大于页面中心或最大阈值,则滑动到下一页(链接),否则滑回其原始位置.例子.我想在我的jquery移动应用程序中应用它.一个示例演示会很棒.谢谢.
有没有办法在没有响应的情况下获取TextInput光标的Y位置(如果可能的话)而没有任何本地(android / ios)黑客攻击?我并不是说光标选择状态的开始和结束。我想要实现的是,当键盘打开时,我想根据光标的位置显示模式,就像在Facebook和Instagram中提到的那样。或者光标可能在哪一行?这样,我也许能够计算光标的位置。
意思是,如果光标在TextInput的开头(屏幕顶部),我想在光标下方显示模式,直到键盘。如果光标位于TextInput的中间或底部(屏幕顶部和Keyboard之间的可见区域的中间),我想在光标上方显示模式。
用例:
而且,由于光标位置和键盘之间有空间,所以我想在光标和键盘之间显示模式。
由于在这种情况下,光标上方到屏幕顶部的空间很大,因此我想在光标上方显示模式。
我正在遵循这两个引用(一个和两个)来创建自定义用户模型,以便通过电子邮件进行身份验证并向其中添加一个额外的字段。
class User(AbstractBaseUser, PermissionsMixin):
email = models.EmailField(
unique=True,
max_length=254,
)
mobile_number = models.IntegerField(unique=True)
is_active = models.BooleanField(default=True)
is_admin = models.BooleanField(default=False)
objects = UserManager()
...
...
class Meta:
db_table = 'auth_user'
...
...
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如您所见,我已经db_table='auth_user'在类的Meta字段中添加了。另外,我在settings.py中包含了AUTH_USER_MODEL = 'accounts.User'和用户模型应用程序(即帐户)INSTALLED_APPS。此外,我从应用程序中删除了Migrations文件夹。
然后尝试迁移:
$ python manage.py makemigrations accounts
Migrations for 'accounts':
accounts/migrations/0001_initial.py:
- Create model User
$ python manage.py migrate accounts
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这给了我一个错误:
django.db.migrations.exceptions.InconsistentMigrationHistory:迁移admin.0001_initial在其依赖项帐户之前应用。数据库'default'上的0001_initial。
如何从现有的django用户模型迁移到自定义用户模型?
假设类Photo用于保存具有选择字段和其他属性的照片:
class Photo(models.Model):
ONLYME = 'O'
FRIENDS = 'F'
PUBLIC = 'P'
CHOICES = (
(ONLYME, "Me"),
(FRIENDS, "Friends"),
(PUBLIC, "Public"),
)
display = models.CharField(max_length=1, choices=CHOICES, blank=True, null=True)
user = models.ForeignKey(User)
description = models.TextField()
pub_date = models.DateTimeField(auto_now=True, auto_now_add=False)
update = models.DateTimeField(auto_now=False, auto_now_add=True)
image = models.ImageField(upload_to=get_upload_file_name, blank=True)
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现在,如何在display属性中将照片的默认值或初始值设置为"朋友" ?
我试图使用Vuex ("^2.1.3")与vuejs ("^2.1.10")这样的项目:
store.js:
import Vue from 'vue';
import Vuex from 'vuex';
Vue.use(Vuex);
export const store = new Vuex.Store({
state: {
inTheaters: [
{
name: 'Resident Evil: The Final Chapter',
poster_url: 'https://blackgirlnerds.com/wp-content/uploads/2017/02/Resident-Evil-The-Final-Chapter-Final-Poster-Featured.jpg',
language: 'English',
},
{
name: 'Irada',
poster_url: 'http://filmywave.com/wp-content/uploads/2017/02/irada-movie-poster-1.jpg',
language: 'Hindi',
},
]
},
});
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main.js:
import store from './store';
new Vue({
router,
components: {App},
template: '<App/>',
store,
}).$mount('#app');
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一些-component.js:
<script>
export default {
name: 'movieListWrapper',
props: {
movieListType: {
type: String,
default: 'in-theateras',
}, …Run Code Online (Sandbox Code Playgroud) 我有一个活动模型,每个活动都有不同的节目.
class Event(models.Model):
title = models.CharField(max_length=200)
class Show(models.Model):
event = models.ForeignKey(Event, on_delete=models.CASCADE)
date_time = models.DateTimeField(unique=True)
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我有另一个Ticket模型.每张票都应该是唯一的.这意味着每张票都是唯一的,并与Show和Seat相关.
class Ticket(models.Model):
show = models.ForeignKey(Show)
seat = models.ForeignKey(Seat)
class Meta:
unique_together = ('show', 'seat')
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我需要根据用户提供的开始日期和结束日期创建节目.假设这是一个JSON帖子:
{
"event_id": 1,
"start_date": "2018-02-16",
"end_date": "2018-02-20",
"time_list": ["11:00 AM", "8:00 PM"]
}
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从上面的JSON示例中,我需要创建Show start,如下所示:
# Start with the start_date as the date, and for each time from the time_list
Show.objects.create(
event = 1,
date_time = datetime.strptime('2018-02-16 11:00 AM', "%Y-%m-%d %I:%M %p")
)
Show.objects.create(
event = 1,
date_time = …Run Code Online (Sandbox Code Playgroud) 我使用amazon s3来存储上传的用户图像.我的问题是:
所以,我的问题是,我该怎么办才能让我网站上的用户上传,下载和删除内容?
在那我有条件:
models.py:
def get_upload_file_name(instance, filename):
return "uploaded_files/%s_%s" %(str(time()).replace('.','_'), filename)
PRIVACY = (
('H','Hide'),
('F','Followers'),
('A','All'),
)
class Status(models.Model):
body = models.TextField(max_length=200)
image = models.ImageField(blank=True, null=True, upload_to=get_upload_file_name)
privacy = models.CharField(max_length=1,choices=PRIVACY, default='F')
pub_date = models.DateTimeField(auto_now_add=True, auto_now=False)
user = models.ForeignKey(User)
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settings.py:
DEFAULT_FILE_STORAGE = 'storages.backends.s3boto.S3BotoStorage'
AWS_ACCESS_KEY_ID = 'FAKEAMAZONKEY'
AWS_SECRET_ACCESS_KEY = 'FAKEAMAZONSECRETKEY'
AWS_STORAGE_BUCKET_NAME = 'fakebucketname'
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更新
用户之间关系的模型
class Person(models.Model):
user = models.OneToOneField(User)
relationships = models.ManyToManyField('self', through='Relationship',
symmetrical=False,
related_name='related_to')
def __unicode__(self):
return self.user.username
def add_relationship(self, person, status):
relationship, created …Run Code Online (Sandbox Code Playgroud) 我正在通过jquery ui docs来使div成为可拖动和可排序的.它确实适用于拖动和排序.但是当我从可拖动的div中取出div时,draggable的宽度会改变,并且不会保留它应该的宽度.
这是演示
JS:
$(function() {
$('#grid-main_content').sortable({
revert:true
});
$('#block-list .block').draggable({
connectToSortable: '#grid-main_content',
helper: 'clone',
revert:'invalid'
});
})
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如果我检查我拖动到sortable的元素,我可以看到一个内联样式被赋予一个带有宽度的可拖动块(通过jquery).
如何保持块已有的样式?谢谢.