小编use*_*531的帖子

'primaryjoin'和'secondaryjoin'如何在SQLAlchemy中为多对多关系工作?

Flask Mega教程中理解一些Flask-SQLAlchemy的东西有些困难.这是代码:

followers = db.Table('followers',
    db.Column('follower_id', db.Integer, db.ForeignKey('user.id')),
    db.Column('followed_id', db.Integer, db.ForeignKey('user.id'))
)

class User(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    nickname = db.Column(db.String(64), unique = True)
    email = db.Column(db.String(120), index = True, unique = True)
    role = db.Column(db.SmallInteger, default = ROLE_USER)
    posts = db.relationship('Post', backref = 'author', lazy = 'dynamic')
    about_me = db.Column(db.String(140))
    last_seen = db.Column(db.DateTime)
    followed = db.relationship('User', 
        secondary = followers, 
        primaryjoin = (followers.c.follower_id == id), 
        secondaryjoin = (followers.c.followed_id == id), 
        backref = db.backref('followers', lazy = 'dynamic'), …
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many-to-many sqlalchemy flask flask-sqlalchemy

16
推荐指数
1
解决办法
1万
查看次数

Flask-Login中使用的"is_authenticated"方法有什么意义?

我正在研究Flask Mega-Tutorial,我遇到了这段代码:

class User(db.Model):
    id = db.Column(db.Integer, primary_key = True)
    nickname = db.Column(db.String(64), unique = True)
    email = db.Column(db.String(120), unique = True)
    role = db.Column(db.SmallInteger, default = ROLE_USER)
    posts = db.relationship('Post', backref = 'author', lazy = 'dynamic')

    def is_authenticated(self):
        return True

    def is_active(self):
        return True

    def is_anonymous(self):
        return False

    def get_id(self):
        return unicode(self.id)

    def __repr__(self):
        return '<User %r>' % (self.nickname)
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is_authenticated,is_active和is_anonymous对我来说似乎很奇怪 - 他们何时会返回除预定义值以外的任何内容?

有人可以向我解释为什么Flask-Login让我使用这些看似无用的方法吗?

python flask web flask-login flask-extensions

14
推荐指数
2
解决办法
6162
查看次数

为什么我不需要在打印之前取消引用C中的字符指针?

为什么这段代码有效?在我打印出来之前,我希望我需要取消引用ptr, printf("%s\n", *ptr);Segmentation Fault如果我尝试这样做,我会得到一个.

#include <stdio.h>

int main(int argc, char *argv[])
{
        char name[] = "Jordan";
        char *ptr = name;
        printf("%s\n", ptr);
}
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希望你们能给我一些见解.

c

6
推荐指数
2
解决办法
7406
查看次数

用箭头在Knockout JS中对表格进行排序

我正在用KnockoutJS创建一个可排序的表。该表工作正常,但是当用户单击该特定标题时,我希望在标题上方显示一个向上箭头。再次单击,向上箭头应朝下。首先,箭头应在“年龄”(Age)列上朝上。

现在,没有箭头出现。是什么导致这种情况在我的代码中发生?

HTML:

<table>
<thead>
<tr data-bind="foreach: headers">
<td data-bind="click: $parent.sort, text: title">
    <span data-bind="if: arrowDown"> v </span>
    <span data-bind="if: arrowUp"> ^ </span>
</td>
</tr>
</thead>
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昏死:

var viewModel = function(){
    var self = this;
    self.people = ko.observableArray([
        {firstName:'James',lastName:'Smith',age:38},
        {firstName:'Susan',lastName:'Smith',age:36},
        {firstName:'Jeremy',lastName:'Smith',age:10},
        {firstName:'Megan',lastName:'Smith',age:7},
        {firstName:'James',lastName:'Jones',age:40},
        {firstName:'Martha',lastName:'Jones',age:36},
        {firstName:'Peggy',lastName:'Jones',age:10}
    ]);
    self.headers = [
        {title:'First Name',sortPropertyName:'firstName', asc:true, arrowDown: false, arrowUp: false},
        {title:'Last Name',sortPropertyName:'lastName', asc:true, arrowDown: false, arrowUp: false},
        {title:'Age',sortPropertyName:'age', asc:true, arrowDown: false, arrowUp: true}
    ];

    self.activeSort = self.headers[2];

    self.sort = function(header, event){
        if(self.activeSort === header) …
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sorting jquery knockout.js

3
推荐指数
1
解决办法
2652
查看次数

悬停时增加透明背景不透明度?

当用户将鼠标悬停在我的div上时,我希望我的背景图像从.1不透明度增加到.5不透明度.

HTML

<div id="list">
    <div class="line_one">om nom nom nom...</div>
    <div class="line_two">18 foods to make you incredibly hungry</div>
</div>
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CSS

#list {
  display:block;
  position: relative; 
  -webkit-transition: all 1s ease; 
  -moz-transition: all 1s ease; 
  -o-transition: all 1s ease; 
  -ms-transition: all 1s ease; 
  transition: all 1s ease;
}

#list::after {
  content: "";
  background: url('test.jpg');
  opacity: 0.1;
  top: 0;
  left: 0;
  bottom: 0;
  right: 0;
  position: absolute;
  z-index: -1;
}
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必须有一种方法可以在没有Javascript的情况下完成此操作.有任何想法吗?

html css opacity

2
推荐指数
1
解决办法
8413
查看次数