How to generate dynamic urls in flask?

Question

I have several records in the database which I Want to form URLs like so:

mysite.com/post/todays-post-will-be-about

The todays-post-will-be-about will be pulled from a Database.

Is there some way I could pull this off in flask?

Asked By: I Love Python

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Source

Answer 1

Flask routes can have parameters as shown here:

@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.

Answered By: Wombatz

Answer 2

Use the @app.route decorator like shown below:

@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

Answered By: sisanared

Answer 3

You can put variable names in your views.py functions. For example:

# 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:

<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:

@app.errorhandler(404)
def not_found_error(error):
    return render_template('404.html', pic=pic), 404

Answered By: ATLUS

Answer 4

I suggest SQLAlchemy http://flask-sqlalchemy.pocoo.org/ It is a very simple and quick way

app.py

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

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

Answered By: Ritero

How to generate dynamic urls in flask?

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