如何通过Playwright Python的page.on("response")获取返回值

原代码中，page.on("response")的回调函数返回值无法直接传递给run函数，因为事件回调是异步触发的。可以通过以下方式实现需求：

方法一：使用可变对象存储响应结果 #

利用列表（可变对象）在回调函数中存储符合条件的响应数据，后续在run函数中读取该数据：

from playwright.sync_api import Playwright, sync_playwright

def run(playwright: Playwright) -> None:
    browser = playwright.chromium.launch(channel="chrome", headless=False)
    context = browser.new_context()
    page = context.new_page()
    
    # 定义可变对象存储响应结果
    response_data = []
    
    def handle_response(response):
        if "api/list" in response.url and response.status == 200:
            # 将结果存入列表
            response_data.append(response.json())
    
    page.on("response", handle_response)
    page.goto("https://www.example.com/")
    
    # 等待页面操作完成，确保响应已被捕获（可根据实际场景调整等待逻辑）
    page.wait_for_load_state("networkidle")
    
    # 使用捕获到的响应数据进行后续操作
    if response_data:
        print("捕获到的响应数据：", response_data[0])
        # 这里添加你的业务逻辑
    
    context.close()
    browser.close()

with sync_playwright() as playwright:
    run(playwright)

方法二：使用page.wait_for_response更精准捕获 #

如果明确要等待某个特定响应，推荐使用page.wait_for_response，它会直接返回匹配的响应对象，无需额外回调：

from playwright.sync_api import Playwright, sync_playwright

def run(playwright: Playwright) -> None:
    browser = playwright.chromium.launch(channel="chrome", headless=False)
    context = browser.new_context()
    page = context.new_page()
    
    # 启动页面操作的同时等待响应
    with page.expect_response(lambda response: "api/list" in response.url and response.status == 200) as response_info:
        page.goto("https://www.example.com/")
    
    # 获取响应数据
    response = response_info.value
    data = response.json()
    
    # 使用响应数据进行后续操作
    print("捕获到的响应数据：", data)
    # 这里添加你的业务逻辑
    
    context.close()
    browser.close()

with sync_playwright() as playwright:
    run(playwright)

关键说明 #

• 方法一适合需要监听多个响应的场景，通过回调收集所有符合条件的结果。
• 方法二更高效，直接等待目标响应，避免不必要的监听，适合明确目标响应的场景。

内容的提问来源于stack exchange，提问作者maruume