python - 使用 Python 条件和 Django 返回 DRY

Question

我要分享的是我项目中的一个常见模式。有些案例比其他案例更长，但模式或多或少是相同的，我想知道如何才能更干。

if status == 'ACTIVE' and p.result != "0":

    messages.add_message(request, messages.ERROR, 
        _('Active Message... bla bla bla'))

    return HttpResponseRedirect(reverse('billing_update'))

if status == 'TOO MANY FAILURES':

    messages.add_message(request, messages.ERROR, 
        _('Failed. Foooobarrrrr'))

    return HttpResponseRedirect(reverse('billing_update'))

if status == 'DEACTIVATED BY MERCHANT':

    messages.add_message(request, messages.ERROR, 
        _('Deactivated derpa derp'))

    return HttpResponseRedirect(reverse('billing_update'))

这里唯一发生的事情是根据状态设置消息。必须有一种方法只调用返回一次，甚至可以收紧条件。

想法？

Answer 1

它是否符合以下内容:

lib 模块

def add_error_message(request, status, result=0):
    if status == 'ACTIVE' and not result:
        msg = 'Active Message... bla bla bla'
    elif status == 'TOO MANY FAILURES':
        msg = 'Failed. Foooobarrrrr'
    elif status == 'DEACTIVATED BY MERCHANT':
        msg = 'Deactivated derpa derp'
    messages.add_message(request, messages.ERROR, _(msg))

如果不是'p.result'条件，字典会好得多:

MESSAGES = {
    'ACTIVE': 'Active Message... bla bla bla',
    'TOO MANY FAILURES': 'Failed. Foooobarrrrr',
    'DEACTIVATED BY MERCHANT': 'Deactivated derpa derp',
}
def add_error_message(request, status):
    try:
        messages.add_message(request, messages.ERROR, _(MESSAGES[status]))
    except KeyError:
        pass

查看模块

from lib import add_error_message

def your_view(request, *args, **kwargs):
    # ...
    add_error_message(request, status, p.result if p else 0)
    return HttpResponseRedirect(reverse('billing_update'))