如何在不使用请求上下文的情况下在烧瓶中呈现模板

how to render template in flask without using request context(如何在不使用请求上下文的情况下在烧瓶中呈现模板)

问题描述

所以我正在为这个项目开发这个烧瓶应用程序，我需要它在定时变量处循环运行，以检查某些变量的状态，然后相应地给出输出.但是，我遇到的问题是我需要在循环重新启动之前在 Flask 中渲染一个模板.在 http://flask.pocoo.org/ 的更新日志中，它表明可以在不使用的情况下渲染模板请求上下文，但我还没有看到任何真实的例子.那么有没有一种方法可以在 Flask 中呈现模板而无需使用请求上下文而不会出现任何错误?感谢您提供任何帮助.

So there's this flask app that I'm working on for this project and I need it to run in a loop at timed variables to check for the status of certain variables and then give a output accordingly. However, the problem I have is I need to render a template in Flask before the loop restarts. In the changelog on http://flask.pocoo.org/ it's indicated that it's possible to render templates without using the request context but I haven't seen any real examples of this. So is there a way to render templates in Flask without having to use the request context without getting any errors? Any help that can be given is appreciated.

更新:这是我正在使用的代码

UPDATE: Here's the code I'm working with

from flask import Flask, render_template, request, flash, redirect, url_for
import flask
import time
from flask.ext.assets import Environment, Bundle
from flask_wtf import Form 
from wtforms import TextField, TextAreaField, SubmitField
from wtforms.validators import InputRequired

CSRF_ENABLED = True
app = Flask(__name__)
app.secret_key = 'development key'
app = flask.Flask('my app')
assets = Environment(app)
assets.url = app.static_url_path
scss = Bundle('scss/app.scss', filters='scss', output='css/app.css')
assets.register('app_scss', scss)

@app.route('/')
def server_1():
    r=1
    g=2
    b=3
    i=g
    if i == g:
        with app.app_context():
            print "Loading Template..."
            rendered = flask.render_template('server_1.html', green=True)
            print "Success! Template was loaded with green server status..."
            time.sleep(5)


if __name__ == '__main__':
    app.run(port=5000, debug=True)

推荐答案

您可以通过将您的应用程序绑定为当前应用程序来实现.然后你可以使用 render_template() 从你的模板目录渲染一个模板，或者 render_template_string() 直接从一个存储在字符串中的模板渲染:

You can do it by binding your application as the current application. Then you can use render_template() to render a template from your template directory, or render_template_string() to render directly from a template stored in a string:

import flask
app = flask.Flask('my app')

with app.app_context():
    context = {'name': 'bob', 'age': 22}
    rendered = flask.render_template('index.html', **context)

with app.app_context():
    template = '{{ name }} is {{ age }} years old.'
    context = {'name': 'bob', 'age': 22}
    rendered = flask.render_template_string(template, **context)

您也可以绕过 Flask 直接进入 Jinja2:

Alternatively you could bypass Flask and go directly to Jinja2:

import jinja2
template = jinja2.Template('{{ name }} is {{ age }} years old.')
rendered = template.render(name='Ginger', age=10)

更新

您可能希望将内容流式传输回发出请求的客户端.如果是这样，您可以编写一个生成器.这样的事情可能会奏效:

It appears that you might be wanting to stream content back to the requesting client. If so you could write a generator. Something like this might work:

import time
from flask import Flask, Response, render_template_string
from flask import stream_with_context

app = Flask(__name__)

@app.route("/")
def server_1():
    def generate_output():
        age = 0
        template = '<p>{{ name }} is {{ age }} seconds old.</p>'
        context = {'name': 'bob'}
        while True:
            context['age'] = age
            yield render_template_string(template, **context)
            time.sleep(5)
            age += 5

    return Response(stream_with_context(generate_output()))

app.run()

这里有一些关于流式传输的文档烧瓶.

Here is some documentation on streaming with Flask.

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