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xmz7891
b62ac865ea
被控端 - 修复 MTU 问题：夹 MSS + 路径 MTU 探测（公网可用率 48% -> 100%） - 复用 HTTP 连接（免掉每次新建 TLS 的 30% 握手失败） - 每分钟回头找局域网，自动切回快 400 倍的路径 - 画面心跳帧：静止时省流量，客户端仍知道流是活的 - 锁屏时给出可操作的提示 手表端 - 内置 GTS Root R4 / GlobalSign Root CA，修 SSLHandshakeException - 连接错误翻译成人话（证书/时间/解析/超时分别提示） - 经中转连上后用服务器报的地址刷新已配对记录 - 新增局域网连接页面、服务器延迟检测 服务器端 - 支持 chunked 请求体（修每 16 秒一次的连接错乱与大面积 502） - 网络调优脚本 net-tune.sh（开机自动应用） - 管理页：批量删除、网络体检、下载区 - 隧道改 HTTP/2 + 双连接 文档 - 许可协议 / 安装前必读 / 安装后须知（作者：小麦子）
232 lines
9.2 KiB
Python
232 lines
9.2 KiB
Python
# -*- coding: utf-8 -*-
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"""配对码被占用时，在电脑上弹通知。
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用户的要求：
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"如果一个设备设置了配对码 789113，另一个设备也想设置 789113，
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那不可以，你就弹通知，说该配对码已被使用，请换一个配对码"
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现状：
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中转服务器**已经**会拒绝（返回 "这个配对码已经被另一台设备用了，换一个"），
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被控端也抛了这个错误 —— 但只是写进日志，用户根本看不到。
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而且主循环会当成网络错误一直重试，刷屏。
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要做的：
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1. 中转返回里加 codeTaken 标记（让客户端能精确识别，不靠匹配中文）
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2. 被控端加 RelayCodeTaken 异常
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3. 加 on_alert 回调 —— 服务器把它接到界面通知上
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4. 主循环遇到冲突时：**弹一次通知就停下来等**，不要反复重试刷屏
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用户换了配对码之后自动继续。
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"""
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from __future__ import annotations
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import sys
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from pathlib import Path
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ROOT = Path(r"C:\Users\shu\Desktop\deepseekagent\LanRemote")
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RELAY_SERVER = ROOT / "relay" / "relay.py"
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RELAY_CLIENT = ROOT / "windows" / "lanremote" / "relay.py"
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SERVER = ROOT / "windows" / "lanremote" / "server.py"
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# ============ 1. 中转：返回加 codeTaken ============
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SRV_OLD = ''' known = self.codes.get(code)
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if known and known != device_id:
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return {"ok": False,
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"error": "这个配对码已经被另一台设备用了，换一个"}'''
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SRV_NEW = ''' known = self.codes.get(code)
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if known and known != device_id:
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# codeTaken 让客户端能精确识别这种冲突，
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# 不用去匹配中文错误文案（文案一改就坏）。
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self.log(f"配对码 {code} 已被 {known} 占用，拒绝 {device_id}")
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return {"ok": False,
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"codeTaken": True,
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"error": "该配对码已被使用，请换一个配对码"}'''
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# ============ 2. 被控端：异常类 ============
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EXC_ANCHOR = '''class _AuthLost(Exception):'''
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EXC_NEW = '''class RelayCodeTaken(RelayError):
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"""配对码被别的设备占了。
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这**不是网络问题**，重试没用 —— 要等用户换一个配对码。
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所以主循环遇到它不能当成普通错误一直重试（会刷屏），
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而是弹一次通知然后停下等。
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"""
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class _AuthLost(Exception):'''
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# ============ 3. 被控端：__init__ 加 on_alert ============
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INIT_OLD = ''' def __init__(self, store, log=None, on_session=None) -> None:
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self.store = store
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self._log = log or (lambda _m: None)
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self._on_session = on_session'''
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INIT_NEW = ''' def __init__(self, store, log=None, on_session=None, on_alert=None) -> None:
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self.store = store
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self._log = log or (lambda _m: None)
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self._on_session = on_session
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# 出需要用户处理的事（比如配对码被占）时调这个，用来弹通知
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self._on_alert = on_alert'''
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INIT2_OLD = ''' self.sessions_served = 0'''
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INIT2_NEW = ''' self.sessions_served = 0
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# 配对码冲突的说明。非空表示现在卡在这，等用户换码。
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self.code_conflict = ""
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self._alerted_code = ""'''
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# ============ 4. 被控端：_register 抛专用异常 ============
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REG_OLD = ''' if not reply.get("ok"):
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raise RelayError(str(reply.get("error") or "登记被拒"))'''
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REG_NEW = ''' if not reply.get("ok"):
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message = str(reply.get("error") or "登记被拒")
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# 配对码被占：精确判断（服务端会带 codeTaken），
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# 兼容老服务端就匹配关键字。
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if reply.get("codeTaken") or "已被" in message and "配对码" in message:
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raise RelayCodeTaken(message)
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raise RelayError(message)
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# 登记成功 -> 之前的冲突解除了
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self.code_conflict = ""'''
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# ============ 5. 被控端：主循环处理冲突 ============
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RUN_OLD = ''' except _AuthLost:
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self._log("中转说令牌失效，重新登记")
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self.token = ""
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time.sleep(1)'''
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RUN_NEW = ''' except _AuthLost:
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self._log("中转说令牌失效，重新登记")
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self.token = ""
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time.sleep(1)
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except RelayCodeTaken as exc:
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# 配对码被别的设备占了。这不是网络问题，重试没用。
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# 弹一次通知，然后**停下来等用户换码**。
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message = str(exc) or "该配对码已被使用，请换一个配对码"
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self.online = False
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self.token = ""
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self.last_error = message
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self.code_conflict = message
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current = self.store.pairing_code()
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if current != self._alerted_code:
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# 只在换了新码之后再弹，避免反复弹同一个码的通知
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self._alerted_code = current
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self._log(f"配对码冲突：{message}")
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if self._on_alert:
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try:
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self._on_alert("配对码被占用", message)
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except Exception: # noqa: BLE001
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pass
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# 等用户换码。每 5 秒看一次配对码变没变。
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for _ in range(12):
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if not self._running:
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return
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time.sleep(0.5)
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if self.store.pairing_code() != current:
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self._log("配对码变了，重新登记")
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self._alerted_code = ""
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break
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continue'''
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# ============ 6. 被控端：status 里带上冲突信息 ============
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STATUS_OLD = ''' "candidates": list(self.candidates),'''
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STATUS_NEW = ''' "candidates": list(self.candidates),
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"codeConflict": self.code_conflict,'''
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# ============ 7. 服务器：把 on_alert 接到界面通知 ============
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WIRE_OLD = ''' self._relay = RelayLink(self.store, log=self._log,
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on_session=self.adopt_relay_socket)'''
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WIRE_NEW = ''' self._relay = RelayLink(self.store, log=self._log,
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on_session=self.adopt_relay_socket,
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on_alert=self._alert)'''
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def patch(path: Path, pairs, label: str) -> bool:
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text = path.read_text(encoding="utf-8")
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ok = True
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for old, new, tag in pairs:
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if old not in text:
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print(f" [{label}] [跳过] {tag} —— 锚点没找到")
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ok = False
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continue
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text = text.replace(old, new, 1)
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print(f" [{label}] {tag}")
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path.write_text(text, encoding="utf-8", newline="\n")
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return ok
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def main() -> int:
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ok = True
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ok &= patch(RELAY_SERVER, [(SRV_OLD, SRV_NEW, "冲突返回加 codeTaken")],
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"中转")
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ok &= patch(RELAY_CLIENT, [
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(EXC_ANCHOR, EXC_NEW, "RelayCodeTaken 异常"),
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(INIT_OLD, INIT_NEW, "__init__ 加 on_alert"),
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(INIT2_OLD, INIT2_NEW, "__init__ 加冲突字段"),
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(REG_OLD, REG_NEW, "_register 抛专用异常"),
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(RUN_OLD, RUN_NEW, "主循环处理冲突"),
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(STATUS_OLD, STATUS_NEW, "status 带冲突信息"),
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], "被控端")
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# 服务器：加 _alert 方法 + 接线
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text = SERVER.read_text(encoding="utf-8")
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if WIRE_OLD in text:
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text = text.replace(WIRE_OLD, WIRE_NEW, 1)
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print(" [服务器] on_alert 接线")
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else:
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print(" [服务器] [跳过] 接线锚点没找到")
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ok = False
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alert_method = ''' def _alert(self, title: str, text: str) -> None:
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"""给用户看的通知（配对码被占这类必须让人知道的事）。
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走界面的通知，用户在电脑上能直接看到。
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界面还没起来时退回写日志，至少不丢信息。
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"""
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self._log(f"[需要处理] {title}：{text}")
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try:
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ui = getattr(self.actions, "ui", None)
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if ui is not None and hasattr(ui, "notify"):
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ui.notify(title, text)
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except Exception: # noqa: BLE001
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pass
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def relay_status(self) -> dict:'''
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if "def _alert(" in text:
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print(" [服务器] 已经有 _alert")
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elif " def relay_status(self) -> dict:" in text:
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text = text.replace(" def relay_status(self) -> dict:",
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alert_method, 1)
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print(" [服务器] 加了 _alert 方法")
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else:
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print(" [服务器] [失败] 找不到 relay_status")
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ok = False
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SERVER.write_text(text, encoding="utf-8", newline="\n")
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import ast
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for label, path in (("中转", RELAY_SERVER), ("被控端", RELAY_CLIENT),
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("服务器", SERVER)):
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try:
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ast.parse(path.read_text(encoding="utf-8"))
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print(f" {label} 语法 OK")
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except SyntaxError as exc:
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print(f" {label} 语法错误: {exc}")
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ok = False
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return 0 if ok else 1
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if __name__ == "__main__":
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sys.exit(main())
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