Python Telegram脚本遇无效用户名报错，如何让代码继续处理下一个用户？

你在编写Telegram脚本时遇到ValueError异常，提示No user has "bvbjjkhjb" as username，希望遇到无效用户名时跳过该用户，继续处理下一个。

原代码的问题在于：当mode=2时，client.get_input_entity(user['username'])这行代码未被包含在try-except块中，抛出的ValueError无法被捕获，直接导致脚本中断。

修改方案 #

将获取receiver的逻辑移到try块内部，同时新增专门的异常分支处理无效用户名场景，确保异常被捕获后跳过当前用户继续执行。

修改后的完整代码 #

for user in users:
    receiver = None
    if mode == 2:
        if user['username'] == "":
            continue
    elif mode == 1:
        receiver = InputPeerUser(user['id'], user['access_hash'])
    else:
        print(re+"[!] Invalid Mode. Exiting.")
        client.disconnect()
        sys.exit()

    try:
        if mode == 2:
            receiver = client.get_input_entity(user['username'])
        
        print(gr+"[+] Sending Message to:", user['name'])
        client.send_message(receiver, data.format(user['name']))
        print(gr+"[+] Waiting {} seconds".format(SLEEP_TIME))
        time.sleep(SLEEP_TIME)
    except PeerFloodError:
        print(re+"[!] Getting Flood Error from telegram. \n[!] Script is stopping now. \n[!] Please try again after some time.")
        client.disconnect()
        sys.exit()
    except ValueError as e:
        print(re+f"[!] Invalid username: {user['username']}, error: {e}")
        print(re+"[!] Skipping this user...")
        continue
    except Exception as e:
        print(re+"[!] Error:", e)
        print(re+"[!] Trying to continue...")
        continue

client.disconnect()
print("Done. Message sent to all users.")

修改说明 #

• 将mode=2下获取receiver的代码移入try块，确保get_input_entity抛出的异常能被捕获。
• 新增except ValueError as e分支，专门处理无效用户名的情况，输出更精准的提示信息。
• 保留原有的PeerFloodError处理逻辑（遇限流直接退出）和其他异常的通用处理（跳过当前用户继续执行）。

内容的提问来源于stack exchange，提问作者Dax Shit