Question

我试图在django项目启动时创建html文件，方法是在project / settings.py中包含如下代码：

def Brander():
    import configparser
    config = configparser.ConfigParser()
    config.read('settings.ini')
    version = config['PROJECT']['version']
    APPNAME = config['BRANDING']['appname']
    APPCOMPANY = config['BRANDING']['appcompany']
    APPCOMPANYLINK = config['BRANDING']['appcompanysite']
    APPLINK = config['BRANDING']['appsite']
    from django.contrib.staticfiles import finders
    filen = finders.find('clinic/brandedfooter.html')
    f = open(filen, "w")
    s = f"""
    <div class="col-lg-8 col-md-8 d-none d-md-block d-lg-block">
        <span class="text-muted float-right"><i>My OP and IP Clinic - <a href="{APPLINK}">{APPCOMPANY} by </a><a href="{APPCOMPANYLINK}">{APPCOMPANY}</a></i></span>
    </div>
    """
    f.write(s)

我的项目/settings.ini包含：

[PROJECT]
version = 0.0.1

[BRANDING]
appname = MyOPIP
appcompany = Droidzone
appcompanysite = https://droidzone.in
appsite = https://myopip.com

当以上代码作为独立的python脚本运行时，一切正常，并且生成了html文件。但是，当将其作为manage.py runserver的一部分执行时，出现以下错误：

joel@hp:~/myappointments$ ./manage.py runserver
Traceback (most recent call last):
File "./manage.py", line 15, in <module>
    execute_from_command_line(sys.argv)
File "/home/joel/.local/lib/python3.6/site-packages/django/core/management/__init__.py", line 381, in execute_from_command_line
    utility.execute()
File "/home/joel/.local/lib/python3.6/site-packages/django/core/management/__init__.py", line 325, in execute
    settings.INSTALLED_APPS
File "/home/joel/.local/lib/python3.6/site-packages/django/conf/__init__.py", line 57, in __getattr__
    self._setup(name)
File "/home/joel/.local/lib/python3.6/site-packages/django/conf/__init__.py", line 44, in _setup
    self._wrapped = Settings(settings_module)
File "/home/joel/.local/lib/python3.6/site-packages/django/conf/__init__.py", line 107, in __init__
    mod = importlib.import_module(self.SETTINGS_MODULE)
File "/usr/lib/python3.6/importlib/__init__.py", line 126, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 994, in _gcd_import
File "<frozen importlib._bootstrap>", line 971, in _find_and_load
File "<frozen importlib._bootstrap>", line 955, in _find_and_load_unlocked
File "<frozen importlib._bootstrap>", line 665, in _load_unlocked
File "<frozen importlib._bootstrap_external>", line 678, in exec_module
File "<frozen importlib._bootstrap>", line 219, in _call_with_frames_removed
File "/home/joel/myappointments/myappointments/settings.py", line 258, in <module>
    Brander()
File "/home/joel/myappointments/myappointments/settings.py", line 16, in Brander
    version = config['PROJECT']['version']
File "/usr/lib/python3.6/configparser.py", line 959, in __getitem__
    raise KeyError(key)
KeyError: 'PROJECT'

我无法理解为什么仅在Django中启动代码时会发生此错误。

Answer 1

您可以使用Django Constance来定义设置。例如：

CONSTANCE_CONFIG = {
    'VERSION': ('0.0.1', 'Version'),
}

然后将'constance.context_processors.config'添加到上下文处理器中，例如上述的documentation。然后在模板中使用它：

{{ config.VERSION }}

在settings.py中运行代码的Django特定错误

1 个答案: