如何通过WSGI实现Flask应用程序按路径分派？ [英] How to implement Flask Application Dispatching by Path with WSGI?

问题描述

我希望将单个域用作多个容器应用程序的分段环境，最终将运行在各自的域中。

类似于：

$ b

• example_staging.com/app1


• example_staging.com/app2


• example_staging.com/app3

其中：

• example_staging.com/app1的行为与app1.example_staging.com相同


• example_staging.com/app2的行为与app2.example_staging.com相同


• example_staging.com/app3的行为与app3.example_staging.com相同

或：

• example_staging.com/app1的行为与app1.com相同


• example_staging.com/app2的行为与app2.com相同


• example_staging.com/app3的行为与app3.com相同

初级应用程序：

  from flask import Flask 
 
 app = Flask（__ name__）
 
 @ app.route（'/'） 
 def hello_wor ld（）：
返回'来自Flask的Hello！'
  

WSGI Starter Config File： / p>

  import sys 
 
 project_home = u'/ home / path / sample1'
 
如果project_home不在sys.path中：
 sys.path = [project_home] + sys.path 
 
从应用程序导入应用程序作为应用程序
  
 
 
 
参考：
 
 
  http://flask.pocoo.org/docs/0.10/patterns/appdispatch/  
 
 
 
我不知道在哪里添加文档中给出的代码作为示例，以及create_app，default_app，get_user_for_prefix应该是什么样子。
注意：使用PythonAnywhere  
 
 
解决方案 
 
 
 Glenns输入后的WSGI配置文件：
 
 
  import sys 
 
＃将项目目录添加到sys.path 
 project_home = u'/ home / path / app1'
如果project_home不在sys.path中：
 sys.path = [project_home] + sys.path 
 
 from werkzeug.wsgi import DispatcherMiddleware 
从应用程序导入应用程序作为app1 $ b $从app2.app导入应用程序as app2 
 from app3.app import app as app3 
 
 application = DispatcherMiddleware（app1，{
'/ app2'：app2，
'/ app3'：app3 
 $）
  
文件夹结构：
 
 
  app1文件夹
 app2文件夹
 app3文件夹
  
 
 
 
 
 
创建每个单独的应用程序，并让他们在各自的域上进行响应。然后创建一个新的应用程序，从各个应用程序中导入应用程序变量，并使用 DispatcherMiddleware 您链接到文档页面上的标题为合并应用程序。
 
I would like to use a single domain as a Staging Environment for multiple flask applications that will eventually run on their own domains.

Something like:


example_staging.com/app1
example_staging.com/app2
example_staging.com/app3


where:


example_staging.com/app1 acts same as app1.example_staging.com
example_staging.com/app2 acts same as app2.example_staging.com
example_staging.com/app3 acts same as app3.example_staging.com


or:


example_staging.com/app1 acts same as app1.com
example_staging.com/app2 acts same as app2.com
example_staging.com/app3 acts same as app3.com


Starter app:
from flask import Flask

app = Flask(__name__)

@app.route('/')
def hello_world():
    return 'Hello from Flask!'
WSGI Starter Config File:
import sys

project_home = u'/home/path/sample1'

if project_home not in sys.path:
    sys.path = [project_home] + sys.path

from app import app as application
Refering to:

http://flask.pocoo.org/docs/0.10/patterns/appdispatch/

I don't know where to add the code given in the documents as an example and what create_app, default_app, get_user_for_prefix should look like.

Note: Using PythonAnywhere

SOLUTION

WSGI Config File after Glenns input:
import sys

# add your project directory to the sys.path
project_home = u'/home/path/app1'
if project_home not in sys.path:
    sys.path = [project_home] + sys.path

from werkzeug.wsgi import DispatcherMiddleware
from app import app as app1
from app2.app import app as app2
from app3.app import app as app3

application = DispatcherMiddleware(app1, {
    '/app2':    app2,
    '/app3':    app3
})
Folder Structure:
app1 folder
    app2 folder
    app3 folder

解决方案
The key thing to note, here, is that you'll actually have 4 apps (3 individual apps and one combined app). This is ignoring the staging/live distinction because staging and live are just copies of each other in different directories. 

Create each of the individual apps and get them responding on their individual domains. Then create a new app that imports the application variables from the individual apps and combines them using the DispatcherMiddleware like the example under the heading "Combining Applications" on the doc page you link to.

                        
这篇关于如何通过WSGI实现Flask应用程序按路径分派？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！