(Python)函数修饰器将剧作家Page对象传递给包装函数

Question

：我正在尝试创建一个函数装饰器，它将一个剧作家Page(playwright.sync_api._generated.Page)对象)传递给包装好的函数。

发出：对于大多数函数调用，我将能够返回函数调用返回的值。但是，由于剧作家需要browser.close()调用，所以我不能简单地返回Page对象。我不确定问题是在(1)我定义的函数装饰器中，还是(2)函数装饰符的用法中。

我试着在pytest固定装置之后给我的装潢师做模特。使用pytest，我会做这样的事情：

@pytest.fixture(scope="module")
def playwright_page():
    with sync_playwright() as play:
        browser = play.chromium.launch()
        page = browser.new_page()
        yield page
        browser.close()

然后

def open_google(playwright_page):
    playwright_page.goto("https://google.com")

函数译码器

>>> from playwright.sync_api import sync_playwright
>>> def playwright_page(func):
        def wrapper(*args, **kwargs):
            with sync_playwright() as play:
                browser = play.chromium.launch()
                page = browser.new_page()
                yield page
                browser.close()
        return wrapper

尝试1

>>> @playwright_page
def open_google():
    page.goto("https://google.com")

    
>>> open_google()
<generator object playwright_page.<locals>.wrapper at 0x0000015C5F6CBC10>

尝试2

>>> @playwright_page
    def open_google():
        page = next(page)
        page.goto("https://google.com")

    
>>> open_google()
<generator object playwright_page.<locals>.wrapper at 0x0000015C5FDB7F90>

Answer 1

我应该调用Page对象并传递Page对象，而不是试图生成pytest对象，比如pytest的对象。

>>> from playwright.sync_api import sync_playwright
>>> def playwright_page(func):
        def wrapper():
            with sync_playwright() as play:
                browser = play.chromium.launch()
                page = browser.new_page()
                results = func(page)
                browser.close()
                return results
        return wrapper


>>> @playwright_page
    def open_google(page):
        page.goto("https://google.com")