小编Tru*_*rby的帖子

简单的angular.js示例有时不会加载

问题:

我开始在我的项目中使用angular.js,在开发过程中,我注意到控制器有时不加载,所以我尝试删除项目的一部分,直到尽可能小的例子,但问题仍然存在.

码:

的index.html

<!DOCTYPE html>
<html lang="en">
<head>
    <title>Test</title>
</head>
<body ng-app="myApp">
    <div ng-controller="TestController">
        <input ng-model="testText" type="text" placeholder="Enter text">
    </div>
    <script src="/static/js/angular.js"></script>
    <script src="/static/js/app.js"></script>
</body>
</html>
Run Code Online (Sandbox Code Playgroud)

app.js

console.log("INIT");
angular.module('myApp', [])
.controller('TestController', ['$scope', function($scope) {
    $scope.testText = '172.17.2.1';
    console.log("SCOPE");
}]);
Run Code Online (Sandbox Code Playgroud)

注意:

"INIT"部分始终显示在控制台中.有时(或大部分时间)Altough"SCOPE"部分不会使输入字段无法填充.

版本:

Chrome:36.0.1985.125

Angular.js:1.3.14

javascript angularjs

7
推荐指数
1
解决办法
1080
查看次数

SQLAlchemy:从特定用户的favorite_series中检索所有剧集

我有用户谁可以有他最喜欢的系列和有系列作为外键的剧集,我试图从最喜欢的用户系列中检索所有剧集.我正在使用Flask-SQLAlchemy.

数据库:

db = SQLAlchemy(app)

# cross table for user-series
favorite_series = db.Table('favorite_series',
    db.Column('user_id', db.Integer, db.ForeignKey('user.id')),
    db.Column('series_id', db.Integer, db.ForeignKey('series.id'))
)

# user
class User(db.Model):
    __tablename__ = 'user'
    id = db.Column(db.Integer, primary_key=True)
    favorite_series = db.relationship('Series', secondary=favorite_series,
        backref=db.backref('users', lazy='dynamic'))

# series
class Series(db.Model):
     __tablename__ = 'series'
    id = db.Column(db.Integer, primary_key=True)

# episode
class Episode(db.Model):
    __tablename__ = 'episode'
    id = db.Column(db.Integer, primary_key=True)
    series_id = db.Column(db.Integer, db.ForeignKey('series.id'))
    series = db.relationship('Series',
        backref=db.backref('episodes', lazy='dynamic'))
Run Code Online (Sandbox Code Playgroud)

朋友用SQL帮助我

select user_id,series.name,episode.name from (favorite_series left join series on favorite_series.series_id = series.id) …
Run Code Online (Sandbox Code Playgroud)

python database many-to-many sqlalchemy flask

2
推荐指数
1
解决办法
3189
查看次数