如何跳过python列表中的错误链接

import requests

# 示例链接列表
links = [
    "https://www.google.com",
    "https://www.invalid-link.com",  # 这是一个无效链接
    "https://www.github.com"
]

# 遍历链接列表
for link in links:
    try:
        # 尝试发送GET请求
        response = requests.get(link)
        # 检查响应状态码
        if response.status_code == 200:
            print(f"Successfully accessed {link}")
        else:
            print(f"Failed to access {link} with status code {response.status_code}")
    except requests.exceptions.RequestException as e:
        # 捕获所有请求异常
        print(f"Error accessing {link}: {e}")

print("Finished processing links.")

import requests

links = [
    "https://www.google.com",
    "https://www.invalid-link.com",
    "https://www.github.com"
]

for link in links:
    try:
        response = requests.get(link)
        response.raise_for_status()  # 如果状态码不是200-299，抛出HTTPError
        print(f"Successfully accessed {link}")
    except requests.exceptions.HTTPError as http_err:
        print(f"HTTP error occurred for {link}: {http_err}")
    except requests.exceptions.ConnectionError as conn_err:
        print(f"Connection error occurred for {link}: {conn_err}")
    except requests.exceptions.Timeout as timeout_err:
        print(f"Timeout error occurred for {link}: {timeout_err}")
    except requests.exceptions.RequestException as req_err:
        print(f"An error occurred for {link}: {req_err}")

print("Finished processing links.")