Question

目标：我正在尝试创建一个函数装饰器，它将 Playwright Page(playwright.sync_api._generated.Page) 对象传递给包装的函数。

问题：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 Playwright 需要 browser.close() 调用，我不能简单地返回 Page 对象。我不确定问题出在 (1) 我定义的函数装饰器，还是 (2) 我对函数装饰器的使用。

我尝试在 pytest 固定装置之后为我的装饰器建模。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")

函数装饰器：

>>> 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 并传递 pytest 对象，而不是尝试产生 func 对象，例如 Page 固定装置。

>>> 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")

(Python) 函数装饰器将 Playwright Page 对象传递给包装函数

1 个答案: