79 lines
2.9 KiB
Python
79 lines
2.9 KiB
Python
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"""Extracted tracking rules with unchanged business behavior."""
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import json
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import random
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import re
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from datetime import datetime, timedelta
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CARD_PREFIX = "__CARD__"
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def card_to_text(content):
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"""把存库的卡片标记转成人话，避免原始 JSON 进入模型上下文"""
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if not content.startswith(CARD_PREFIX):
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return content
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try:
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card = json.loads(content[len(CARD_PREFIX):])
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return f"（已为运单 {card.get('no', '')} 查询过轨迹，当前状态：{card.get('status', '')}）"
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except Exception:
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return "（已为该运单查询过轨迹）"
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def is_tracking_no(text):
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"""判断一句话是不是"整句运单号"
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规则（逐条收紧，避免误判）：
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1. 去掉空格和横杠后，长度 8~20 位，且只能由字母数字组成
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2. 至少包含 3 个数字（英文句子通常只有 1 个数字）
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3. 字母数量不能多于数字（"i want to go 1t how money" 这类会被排除）
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纯数字单号（如 15252125）全部满足上述条件。
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"""
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# 日期格式（如 2026-10-23、2026/10/23）不算运单号
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if re.fullmatch(r"\d{4}[-/.]\d{1,2}[-/.]\d{1,2}", text.strip()):
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return False
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t = re.sub(r"[\s\-]", "", text)
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if not (8 <= len(t) <= 20):
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return False
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if not re.fullmatch(r"[A-Za-z0-9]+", t):
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return False
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digits = sum(c.isdigit() for c in t)
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letters = sum(c.isalpha() for c in t)
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if digits < 3:
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return False
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if letters > digits:
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return False
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return True
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def extract_tracking_no(text):
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"""从一句话里提取运单号：先看整句是不是，再从句子里找最长的字母数字串"""
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t = text.strip()
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if is_tracking_no(t):
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return t
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# 找句子里最长的 8~20 位字母数字连续串
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candidates = re.findall(r"[A-Za-z0-9]{8,20}", t)
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best = None
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for c in candidates:
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if is_tracking_no(c):
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if best is None or len(c) > len(best):
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best = c
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return best
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def build_track(no):
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"""生成固定的模拟轨迹：同一个单号每次查都一样"""
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rnd = random.Random(no)
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steps = [
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("已揽收", "义乌仓"),
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("出口清关放行", "义乌/宁波口岸"),
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("已发车", "中欧班列/卡航 · 阿拉山口"),
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("到达欧洲中转仓", "波兰 · 华沙"),
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("欧盟清关放行", "德国 · 杜伊斯堡"),
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("派送中", "收件人所在城市"),
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]
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gaps = [rnd.randint(6, 40) for _ in steps] # 相邻节点间隔（小时）
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t = datetime.now() - timedelta(hours=sum(gaps))
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nodes = []
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for (text, place), gap in zip(steps, gaps):
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t = t + timedelta(hours=gap)
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nodes.append({"time": t.strftime("%Y-%m-%d %H:%M"),
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"text": text, "place": place})
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nodes.reverse() # 最新的排最上面
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return {"no": no, "status": nodes[0]["text"], "nodes": nodes}
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