Python 如何在烧瓶中生成动态网址？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35107885/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to generate dynamic urls in flask?
提问by I Love Python
I have several records in the database which I Want to form URLs like so:
我在数据库中有几条记录，我想像这样形成 URL：
mysite.com/post/todays-post-will-be-about
mysite.com/post/todays-post-will-be-about
The todays-post-will-be-aboutwill be pulled from a Database.
该todays-post-will-be-about会从数据库中拉出。
Is there some way I could pull this off in flask?
有什么办法可以把它放在烧瓶里吗？
采纳答案by ATLUS
You can put variable names in your views.py functions. For example:
您可以在 views.py 函数中放置变量名称。例如：
# you can also use a particular data type such as int,str
# @app.route('post/<int:id>', methods=['GET', 'POST'])
@app.route('post/<variable>', methods=['GET'])
def daily_post(variable):
#do your code here
return render_template("template.html",para1=meter1, para2=meter2)
To get your database information to display on your site, you'll want to pass parameters into the template. So, in your template you'll reference those parameters like:
为了让您的数据库信息显示在您的站点上，您需要将参数传递到模板中。因此，在您的模板中，您将引用这些参数，例如：
<td>Post Author: {{ para1.author }}</td>
<td>Post Body: {{ para1.body }}</td>
<td>Date Posted: [{{ para2 }}] times</td>
Then when you visit mysite.com/post/anything_here, the 'anything_here' will go into your function and be evaluated as necessary. You'll probably also want to set up 404 page handling, in case someone tries to enter a post manually:
然后，当您访问 mysite.com/post/anything_here 时，“anything_here”将进入您的函数并根据需要进行评估。您可能还想设置 404 页面处理，以防有人尝试手动输入帖子：
@app.errorhandler(404)
def not_found_error(error):
return render_template('404.html', pic=pic), 404
回答by Wombatz
Flask routes can have parameters as shown here:
瓶路线可以有参数显示在这里：
@app.route("post/<identifier>")
def post(identifier): # parameter name must match dynamic route parameter name
the_post = get_from_database_by(identifier)
response = make_response_from_entity(the_post)
return response
How you get the post from the database and how you make a response from that is up to you.
您如何从数据库中获取帖子以及如何从中做出回应取决于您。
回答by sisanared
Use the @app.routedecorator like shown below:
使用@app.route如下所示的装饰器：
@app.route('/post/<post_title>')
def show_post(post_title):
#use post title to fetch the record from db
More examples are available under the Routing section: http://flask.pocoo.org/docs/0.10/quickstart/#routing
路由部分提供了更多示例：http: //flask.pocoo.org/docs/0.10/quickstart/#routing
回答by Ritero
I suggest SQLAlchemy http://flask-sqlalchemy.pocoo.org/It is a very simple and quick way
我建议使用 SQLAlchemy http://flask-sqlalchemy.pocoo.org/这是一个非常简单快捷的方法
app.py
应用程序
from flask import Flask, render_template
try:
from .alchemy import Post, db
except:
from alchemy import Post, db
app = Flask(__name__)
@app.route('/post/<url>')
def post(url):
url = Post.query.filter_by(url=url).first_or_404()
id = url.id
author = url.author
title = url.title
body = url.body
date = url.date
return render_template('post.html', title=title, id=id, author=author, body=body, date=date)
if __name__ == '__main__':
app.run(debug=True)
alchemy.py
炼金术.py
from flask import Flask
from flask.ext.sqlalchemy import SQLAlchemy
import datetime
app = Flask(__name__)
SQLALCHEMY_TRACK_MODIFICATIONS = True
app.config['SQLALCHEMY_DATABASE_URI'] = 'mysql+pymysql://root:pasword@localhost/base'
db = SQLAlchemy(app)
class Post(db.Model):
__tablename__ = "table"
id = db.Column(db.Integer, primary_key=True)
title = db.Column(db.String(200))
url = db.Column(db.String(220))
author= db.Column(db.String(50))
body = db.Column(db.String(50000))
date = db.Column(db.DateTime)
def __init__(self, title, url, author, body):
self.title = title
self.url = url
self.author= author
self.body = body
self.date = datetime.datetime.utcnow()
def __repr__(self):
return '<Post %r>' % self.url
