FlaskApp 使用 mod_wsgi 在 apache 中返回 http 500

Question

我正在尝试通过 apache 和 mod_wsgi 托管我的 python 3.4 烧瓶应用程序。通过烧瓶自己的服务器运行应用程序工作正常。该应用是在虚拟环境 pyvenv-3.4 中制作的。

但是，当尝试在浏览器中连接到 apache 服务器时，它会引发 500 http 错误。附加配置和日志。我认为这与使用 pyvenv 而不是 virtualenv（来自 pip）有关。 Flask 文档告诉我使用这一行来激活虚拟环境

activate_this = '/path/to/env/bin/activate_this.py'

但是，由于文件不存在，这会产生 IOError。我尝试将其指向“激活”文件，然后激活.csh、激活.fish，但没有成功。所有文件都会在 deactivate 行产生 SyntaxError。

如何使用我的 virtualenv 通过 Apache 运行这个应用程序？

flaskapp.wsgi

#!/usr/bin/python
activate_this = '/var/www/FlaskApp/FlaskApp/bin/activate'
execfile(activate_this, dict(__file__=activate_this))
import sys
import logging

logging.basicConfig(stream=sys.stderr)
sys.path.insert(0,"/var/www/FlaskApp/")

from FlaskApp import app as application
application.secret_key = 'some secret key'

Apache 虚拟主机

<VirtualHost *:80>
            ServerName example.org # my server name
            ServerAlias gallifrey 192.168.0.84
            ServerAdmin admin@example.org # my admin
            WSGIScriptAlias /flask /var/www/FlaskApp/flaskapp.wsgi
            <Directory /var/www/FlaskApp/FlaskApp/>
                    Order allow,deny
                    Allow from all
            </Directory>
            Alias /static /var/www/FlaskApp/FlaskApp/static
            <Directory /var/www/FlaskApp/FlaskApp/static/>
                    Order allow,deny
                    Allow from all
            </Directory>
            ErrorLog ${APACHE_LOG_DIR}/error.log
            LogLevel warn
            CustomLog ${APACHE_LOG_DIR}/access.log combined
</VirtualHost>

层次结构

/var/www/FlaskApp
  flaskapp.wsgi
  FlaskApp/
    bin/
      activate
      activate.csh
      activate.fish
      easy_install, easy_install-3.4
      pip, pip3, pip3.4
      python, python3, python3.4
    include/
    lib/
    scripts/
    static/
    templates/
    app.py
    __init__.py

尝试打开网页时出现 http 500 错误：

Apache 错误日志

[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81] mod_wsgi (pid=31629): Target WSGI script '/var/www/FlaskApp/flaskapp.wsgi' cannot be loaded as Python module.
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81] mod_wsgi (pid=31629): Exception occurred processing WSGI script '/var/www/FlaskApp/flaskapp.wsgi'.
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81] Traceback (most recent call last):
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81]   File "/var/www/FlaskApp/flaskapp.wsgi", line 3, in <module>
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81]     execfile(activate_this, dict(__file__=activate_this))
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81]   File "/var/www/FlaskApp/FlaskApp/bin/activate", line 4
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81]     deactivate () {
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81]                   ^
[Fri May 02 10:22:58 2014] [error] [client 192.168.0.81] SyntaxError: invalid syntax

Answer 1

您可以遵循VirtualEnvironments instructions，也可以复制virtualenv 的activate_this.py 脚本的作用：

import sys
import os

old_os_path = os.environ['PATH']
os.environ['PATH'] = os.path.dirname(os.path.abspath(__file__)) + os.pathsep + old_os_path
base = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
if sys.platform == 'win32':
    site_packages = os.path.join(base, 'Lib', 'site-packages')
else:
    site_packages = os.path.join(base, 'lib', 'python%s' % sys.version[:3], 'site-packages')
prev_sys_path = list(sys.path)
import site
site.addsitedir(site_packages)
sys.real_prefix = sys.prefix
sys.prefix = base
# Move the added items to the front of the path:
new_sys_path = []
for item in list(sys.path):
    if item not in prev_sys_path:
        new_sys_path.append(item)
        sys.path.remove(item)
sys.path[:0] = new_sys_path

你可以把它变成一个更通用的函数：

import sys
import os

def activate_venv(path):
    if sys.platform == 'win32':
        bin_dir = os.path.join(path, 'Scripts')
        site_packages = os.path.join(base, 'Lib', 'site-packages')
    else:
        bin_dir = os.path.join(path, 'bin')
        site_packages = os.path.join(BASE, 'lib', 'python%s' % sys.version[:3], 'site-packages')
    os.environ['PATH'] = bin_dir + os.pathsep + os.environ['PATH']
    prev_sys_path = list(sys.path)
    import site
    site.addsitedir(site_packages)
    sys.prefix, sys.real_prefix = path, sys.prefix

    # Move the added items to the front of the path:
    new_sys_path = []
    for item in list(sys.path):
        if item not in prev_sys_path:
            new_sys_path.append(item)
            sys.path.remove(item)
    sys.path[:0] = new_sys_path

将其放入默认 Python 模块搜索路径的模块中，导入 activate_venv 并传入 os.path.dirname(os.path.abspath(__file__)) 的结果：

from somemodule import activate_venv
import os.path
activate_venv(os.path.dirname(os.path.abspath(__file__)))

Answer 2

你试过了吗：

WSGIPythonHome /var/www/FlaskApp

只是让 mod_wsgi 处理设置它还是你？

Answer 3

我尝试在 /etc/httpd/conf/httpd.conf 的上方和外部添加它并且它有效。

WSGIPythonPath /your_virtualenv/lib/python2.7/site-package

Answer 4

当您使用 Python 虚拟环境以及 Django 或 Flask 等应用程序时，您必须确保 Apache 获取 WSGI 路径的位置。

如果你有一个 Debian 发行版并使用包管理器来安装 Apache，你应该有这样的配置文件目录结构： /etc/apache2/.

编辑“/etc/apache2/sites-available/AppName.conf”下的 VirtualHost 配置文件（例如 AppName.conf），并按照以下示例在 Python v3.5 下创建 Flask 应用：

WSGIPythonPath /<PATH_OF_PYTHON_VIRTUAL_ENVIRONMENT>/lib/python3.5/site-packages
<VirtualHost *:80>
        ServerName <HOST or IP_ADDRESS>
        ServerAdmin your-email@domain.com
        WSGIScriptAlias / /var/www/app_name/your_wsgi_app.wsgi
        WSGIPassAuthorization On
        <Directory /var/www/app_name/AppName/>
                Order allow,deny
                Allow from all
        </Directory>
        Alias /static /var/www/app_name/AppName/static
        <Directory /var/www/app_name/AppName/static/>
                Order allow,deny
                Allow from all
        </Directory>

        ErrorLog ${APACHE_LOG_DIR}/error.log
        # e.g. LogLevel debug
        LogLevel <info|warn|error|debug>
        CustomLog ${APACHE_LOG_DIR}/access.log combined
</VirtualHost>