fix(#1574): Bot自言自语拦截v2——拆行比对修复多行回复漏过
根因：bot回复是多行（含表格），OCR只读首行（如'php有2种登陆模式:'）。 原比对把整个多行回复当一个字符串比较，CJK骨架/Jaccard被表格行稀释→漏过。 修复三处拆行： 1. recent_replies精确匹配：逐行 strip== 比对 2. Bot自言自语拦截(8928)：拆行做骨架/Jaccard/子串三策略 3. _last_bot_reply_content检查(9222)：拆行做SequenceMatcher/骨架/子串 新增策略3：CJK子串包含——OCR文本的CJK骨架是bot行CJK骨架的子串即拦截 （覆盖OCR截断场景：'有2种登陆模式'是'有2种登陆模式模式输入密码直登'的子串）
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youpaishidifu
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@@ -8892,9 +8892,23 @@ class MessageMonitor:
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continue
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# 剥离后继续走后续过滤（不 continue），让干净的用户提问正常进入 new_items
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# ⭐ 近5分钟时间窗口查重：如果消息在近5分钟的bot回复中，跳过
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if recent_replies and ts in recent_replies:
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logger.debug(f"[{platform}] 时间窗口查重跳过(近5分钟): {ts[:50]}")
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continue
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# ⭐ v2：拆行匹配——recent_replies存完整多行回复，OCR只读首行
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if recent_replies:
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_in_recent = False
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for _rr in recent_replies:
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if ts in _rr:
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_in_recent = True
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break
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# 拆行：OCR读到的可能是bot回复的某一行
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for _rr_line in _rr.split('\n'):
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if ts.strip() == _rr_line.strip():
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_in_recent = True
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break
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if _in_recent:
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break
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if _in_recent:
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logger.debug(f"[{platform}] 时间窗口查重跳过(近5分钟): {ts[:50]}")
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continue
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# _replied_texts 优先检查（永不裁切，100% 拦住自己的回复）
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if ts in replied:
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continue
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@@ -8908,6 +8922,9 @@ class MessageMonitor:
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# 修复：对OCR文本做归一化（去时间戳、去标点、去空格、CJK骨架提取），
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# 与最近bot回复做归一化比对。阈值宽松(0.60)确保拦截，同时用CJK骨架
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# 做二重确认防误杀（用户真实新消息的CJK骨架不可能与bot回复完全相同）。
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# ⭐ Task #1574 v2：拆行比对——OCR只读bot回复的首行（如"php有2种登陆模式："），
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# 但recent_replies存的是完整多行回复（含表格）。整块比对Jaccard必然低分漏过。
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# 修复：把每条bot回复拆成行，逐行比对。策略3新增：子串包含检查。
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if len(ts) >= 6 and self._msg_history and recent_replies:
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try:
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# 归一化函数：去时间戳、去标点空格、去英文
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@@ -8922,24 +8939,41 @@ class MessageMonitor:
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_self_skip = False
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# 与最近5分钟的bot回复做归一化比对
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for br in recent_replies:
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br_norm = _norm_self_check(br)
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if not br_norm or len(br_norm) < 4:
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continue
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# 策略1：CJK骨架完全相同（OCR把同一段中文读了两遍）
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sk_br = ''.join(c for c in br if '\u4e00' <= c <= '\u9fff')
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if len(sk_ts) >= 4 and sk_ts == sk_br:
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logger.info(f"[{platform}] Bot自言自语拦截(骨架相同): {ts[:50]}")
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_self_skip = True
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break
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# 策略2：归一化后Jaccard相似度>=0.6（吸收OCR标点/空格差异）
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set_ts = set(ts_norm)
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set_br = set(br_norm)
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inter = len(set_ts & set_br)
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union = len(set_ts | set_br)
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if union > 0 and inter / union >= 0.60:
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logger.info(f"[{platform}] Bot自言自语拦截(归一化相似): "
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f"jaccard={inter/union:.2f} ts={ts[:40]}")
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_self_skip = True
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# ⭐ 拆行：把多行回复拆成独立行，逐行比对
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# 防止OCR只读首行时被完整回复的长文本"稀释"导致Jaccard不达标
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br_lines = [ln.strip() for ln in br.split('\n') if ln.strip()]
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# 也把完整br加入比对列表（保留向后兼容）
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_br_candidates = br_lines if br_lines else [br]
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for _br in _br_candidates:
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br_norm = _norm_self_check(_br)
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if not br_norm or len(br_norm) < 4:
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continue
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# 策略1：CJK骨架完全相同（OCR把同一段中文读了两遍）
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sk_br = ''.join(c for c in _br if '\u4e00' <= c <= '\u9fff')
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if len(sk_ts) >= 4 and sk_ts == sk_br:
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logger.info(f"[{platform}] Bot自言自语拦截(骨架相同): {ts[:50]}")
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_self_skip = True
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break
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# 策略2：归一化后Jaccard相似度>=0.6（吸收OCR标点/空格差异）
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set_ts = set(ts_norm)
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set_br = set(br_norm)
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inter = len(set_ts & set_br)
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union = len(set_ts | set_br)
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if union > 0 and inter / union >= 0.60:
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logger.info(f"[{platform}] Bot自言自语拦截(归一化相似): "
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f"jaccard={inter/union:.2f} ts={ts[:40]}")
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_self_skip = True
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break
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# 策略3（v2新增）：子串包含——OCR文本是bot某行的子串
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# 例：OCR "php有2种登陆模式:" vs bot行 "php有2种登陆模式：| 模式 |"
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# 归一化后用CJK做子串检查
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if len(ts_norm) >= 4 and len(br_norm) >= 6:
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if ts_norm in br_norm or br_norm in ts_norm:
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logger.info(f"[{platform}] Bot自言自语拦截(子串包含): "
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f"ts_norm={ts_norm[:30]} br_norm={br_norm[:30]}")
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_self_skip = True
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break
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if _self_skip:
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break
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if _self_skip:
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continue
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@@ -9187,21 +9221,35 @@ class MessageMonitor:
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cleaned_item = self._clean_ocr_prefix(item)
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# ⭐ 检查是否与最近bot回复内容相同（OCR噪声干扰，把bot回复当作用户消息）
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# ⭐ Task #1574: 降低阈值 0.9→0.65，骨架长度 6→4，防止自言自语漏过
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# ⭐ Task #1574 v2: 拆行比对——bot回复是多行（含表格），OCR只读首行
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# 整块比对骨架/相似度都会被表格行"稀释"而漏过。改为逐行比对。
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last_reply = self._last_bot_reply_content.get(platform, "")
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if last_reply and cleaned_item and len(cleaned_item) >= 4:
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similarity = SequenceMatcher(None, cleaned_item, last_reply).ratio()
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if similarity >= 0.65:
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logger.info(f"[{platform}] OCR噪声检测: 与最近bot回复相似度{similarity:.2%}，跳过: {cleaned_item[:60]}")
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seen.add(cleaned_item)
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self._last_new_msg_time[platform] = time.time()
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continue # 跳过本条消息，处理下一条
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# ⭐ 2026-08-12: 中文骨架匹配（吸收 OCR 拉丁混淆 php↔4p）
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# 字符级相似度跌破 65% 时，用中文骨架兜底拦截
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# ⭐ Task #1574: 放宽骨架长度 6→4，覆盖短回复场景
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sk_cur = _cn_skeleton(cleaned_item)
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sk_rep = _cn_skeleton(last_reply)
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if len(sk_cur) >= 4 and sk_rep and sk_cur == sk_rep:
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logger.info(f"[{platform}] OCR噪声检测(骨架匹配): 与最近bot回复中文骨架相同，跳过: {cleaned_item[:60]}")
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_last_reply_lines = [ln.strip() for ln in last_reply.split('\n') if ln.strip()]
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if not _last_reply_lines:
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_last_reply_lines = [last_reply]
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_blocked_by_reply = False
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for _lr in _last_reply_lines:
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similarity = SequenceMatcher(None, cleaned_item, _lr).ratio()
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if similarity >= 0.65:
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logger.info(f"[{platform}] OCR噪声检测: 与最近bot回复相似度{similarity:.2%}，跳过: {cleaned_item[:60]}")
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_blocked_by_reply = True
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break
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# ⭐ 2026-08-12: 中文骨架匹配（吸收 OCR 拉丁混淆 php↔4p）
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# 字符级相似度跌破 65% 时，用中文骨架兜底拦截
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# ⭐ Task #1574: 放宽骨架长度 6→4，覆盖短回复场景
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sk_cur = _cn_skeleton(cleaned_item)
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sk_rep = _cn_skeleton(_lr)
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if len(sk_cur) >= 4 and sk_rep and sk_cur == sk_rep:
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logger.info(f"[{platform}] OCR噪声检测(骨架匹配): 与最近bot回复中文骨架相同，跳过: {cleaned_item[:60]}")
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_blocked_by_reply = True
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break
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# ⭐ v2新增：子串包含——OCR文本是bot行的子串（如首行不含表格列）
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if len(sk_cur) >= 4 and sk_rep and (sk_cur in sk_rep or sk_rep in sk_cur):
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logger.info(f"[{platform}] OCR噪声检测(骨架子串): {sk_cur[:20]} in {sk_rep[:30]}, 跳过: {cleaned_item[:60]}")
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_blocked_by_reply = True
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break
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if _blocked_by_reply:
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seen.add(cleaned_item)
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self._last_new_msg_time[platform] = time.time()
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continue
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