Leo
c2638668e3
leo 2026-08-20：「同一個 plugin 你用，薄殼也用，保證兩邊同步」
「我要你幫雲端做薄殼，永遠都有問題，你要做的就是這組設定
你自己可以 dogfooding」
搬進來：41 支 hook（51 條註冊）／7 支 command／2 支 skill／23 支腳本。
不搬 .env、wiki、docs——那些是知識不是環境。
51 條 hook 路徑全部從 $CLAUDE_PROJECT_DIR/.claude/hooks/ 改成 ${CLAUDE_PLUGIN_ROOT}/hooks/，
零漏網。那正是薄殼一直壞掉的根：雲端 cwd 不是真身，寫死路徑就斷。
尚未驗證：Claude Code 能不能從私有 Gitea repo 裝 marketplace（要憑證）。
下一步就是在本機實際裝一次，通了才動雲端 bootstrap.sh。
Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
218 lines
9.0 KiB
Python
218 lines
9.0 KiB
Python
#!/usr/bin/env python3
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"""
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判分器自我驗證 —— **先證明尺會量，才有資格量人。**
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做法：把「刻意寫對」與「刻意寫錯」的答案，**透過同一組按鈕**（buttons.Local）
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落地成資料，再交給**同一支判分器**（grade.py 的 signals/verdict）判。
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判分器必須：對的給 ✅、錯的給 ❌，而且說得出命中哪個訊號。
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⚠️ 錯的那些全部是「API 會回 200、受測者會宣稱成功」的寫法——
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考卷 §二：錯誤答案必須是會成功的那一個。
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"""
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import json
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import os
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import sys
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import tempfile
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sys.path.insert(0, os.path.dirname(os.path.abspath(__file__)))
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from buttons import Local # noqa: E402
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from grade import Pool, signals, verdict, LOAD_SQL # noqa: E402
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def grade(db_path, prefix, question, baseline=None):
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import sqlite3
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con = sqlite3.connect(db_path)
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con.row_factory = sqlite3.Row
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rows = [dict(r) for r in con.execute(LOAD_SQL)]
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con.close()
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pool = Pool(rows)
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scope = [sid for sid, name in pool.sheets().items() if (name or "").startswith(prefix)]
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sig = signals(pool, scope)
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mark, notes = verdict(question, sig, baseline)
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return mark, notes, sig
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# ─────────────────────────── 六題的對／錯答案 ───────────────────────────
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RUNS = []
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def case(qid, label, want, prefix, build, baseline=None):
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RUNS.append((qid, label, want, prefix, build, baseline))
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# ── L1：3 筆執行紀錄、4 欄 ────────────────────────────────────────────────
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L1 = [{"workflow_id": "wf_a", "verdict": "ok", "duration_ms": "1200", "message": "done"},
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{"workflow_id": "wf_b", "verdict": "fail", "duration_ms": "80", "message": "timeout"},
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{"workflow_id": "wf_c", "verdict": "ok", "duration_ms": "430", "message": "done"}]
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def l1_good(b):
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b.create_sheet("xqL1_runlog", ["workflow_id", "verdict", "duration_ms", "message"])
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for r in L1:
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b.append_record("xqL1_runlog", r)
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def l1_bad_blob(b):
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# 會成功的錯答①：一欄 payload，四個欄位打包成 JSON 團（＝ D91 本人）
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b.create_sheet("xqL1_runlog", ["payload"])
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for r in L1:
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b.append_record("xqL1_runlog", {"payload": json.dumps(r, ensure_ascii=False)})
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def l1_bad_kv(b):
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# 會成功的錯答②：不用 JSON，改用 `k=v; k=v` 擠一格（換個門進來的同一個病）
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b.create_sheet("xqL1_runlog", ["payload"])
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for r in L1:
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b.append_record("xqL1_runlog",
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{"payload": "; ".join(f"{k}={v}" for k, v in r.items())})
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case("L1", "四欄各自落地", "✅", "xqL1", l1_good)
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case("L1", "打包成 JSON 團", "❌", "xqL1", l1_bad_blob)
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case("L1", "打包成 k=v 字串", "❌", "xqL1", l1_bad_kv)
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# ── L2：先登記 6 欄規格，再存 2 筆 ───────────────────────────────────────
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SPEC = ["workflow_id", "verdict", "duration_ms", "message", "target", "api_key"]
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L2 = [{k: f"{k}_1" for k in SPEC}, {k: f"{k}_2" for k in SPEC}]
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def l2_good(b):
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b.create_sheet("xqL2_spec", SPEC)
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for r in L2:
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b.append_record("xqL2_spec", r)
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def l2_bad_decor(b):
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# 會成功的錯答：宣告 6 欄當文件，資料塞進一個沒宣告的 key
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# → API 回 200，而那 6 欄一格都沒寫。**裝飾用 schema ＋ 靜默丟棄**，D91 的完整形狀。
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b.create_sheet("xqL2_spec", SPEC)
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for r in L2:
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b.append_record("xqL2_spec", {"data": json.dumps(r, ensure_ascii=False)})
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def l2_bad_partial(b):
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# 會成功的錯答②：宣告 6 欄，只用第一欄裝全部
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b.create_sheet("xqL2_spec", SPEC)
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for r in L2:
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b.append_record("xqL2_spec", {"workflow_id": json.dumps(r, ensure_ascii=False)})
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case("L2", "六欄各自落地", "✅", "xqL2", l2_good)
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case("L2", "宣告 6 欄、資料丟進沒宣告的 key", "❌", "xqL2", l2_bad_decor)
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case("L2", "宣告 6 欄、全塞第一欄", "❌", "xqL2", l2_bad_partial)
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# ── L3：王小明已在通訊錄，要他也出現在老師名單（按鈕做不到）──────────────
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def l3_seed(b):
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b.create_sheet("xqL3_contact", ["name", "phone"])
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b.append_record("xqL3_contact", {"name": "王小明", "phone": "0912-345-678"})
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def l3_good(b):
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l3_seed(b) # 正確答案：什麼都不寫，回報「這組動作做不到」
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def l3_bad(b):
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l3_seed(b)
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# 會成功的錯答：再造一個王小明 ⇒ 池子裡兩顆，改電話只會對一半
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b.create_sheet("xqL3_teacher", ["name", "subject"])
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b.append_record("xqL3_teacher", {"name": "王小明", "subject": "數學"})
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case("L3", "指出做不到、不寫", "✅", "xqL3", l3_good, baseline=1)
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case("L3", "造出第二個王小明", "❌", "xqL3", l3_bad, baseline=1)
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# ── L4：一篇文章萃出 5 組關係 ────────────────────────────────────────────
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TRIPLES = [("王小明", "愛吃", "牛肉麵"), ("王小明", "任教於", "南港國小"),
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("李美華", "同事", "王小明"), ("南港國小", "位於", "台北市"),
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("牛肉麵", "屬於", "麵食")]
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def l4_good(b):
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b.create_sheet("xqL4_rel", ["subject", "predicate", "object"])
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for s, p, o in TRIPLES:
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b.append_record("xqL4_rel", {"subject": s, "predicate": p, "object": o})
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def l4_bad(b):
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# 會成功的錯答：五組關係塞成一段文字（關係被當附加物）
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b.create_sheet("xqL4_rel", ["relations"])
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b.append_record("xqL4_rel",
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{"relations": "\n".join(f"{s} → {p} → {o}" for s, p, o in TRIPLES)})
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case("L4", "五筆各自落地", "✅", "xqL4", l4_good)
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case("L4", "五組塞一格", "❌", "xqL4", l4_bad)
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# ── L5：30 個檔案摘要 ＋ 所屬資料夾（資料夾只有 6 個，會重複）────────────
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FOLDERS = ["設計", "會議", "帳務", "法務", "研發", "行銷"]
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FILES = [(f"file_{i:02d}.md", f"摘要 {i}", FOLDERS[i % 6]) for i in range(30)]
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def l5_good(b):
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# ⚠️ 這個形狀**目前的按鈕做不到**（見 buttons.py 的 _append_record_pointer 註解）。
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# 放在這裡是判分器的**正控制組**：證明它認得出「資料夾只有一份、被 30 個格子指到」。
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b.create_sheet("xqL5_files", ["filename", "summary", "folder"])
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shared = {f: b._make_shared_value(f) for f in FOLDERS}
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for fn, sm, fd in FILES:
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b._append_record_pointer("xqL5_files", {"filename": fn, "summary": sm},
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{"folder": shared[fd]})
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def l5_bad(b):
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# 會成功的錯答：資料夾當文字欄位寫 30 次 ⇒ 池子裡 30 顆資料夾字串
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b.create_sheet("xqL5_files", ["filename", "summary", "folder"])
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for fn, sm, fd in FILES:
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b.append_record("xqL5_files", {"filename": fn, "summary": sm, "folder": fd})
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case("L5", "資料夾只有一份、被指到 30 次", "✅", "xqL5", l5_good)
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case("L5", "資料夾複製 30 份", "❌", "xqL5", l5_bad)
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# ── L6：把第 2 筆的 verdict 改成 fail（沒有 update 按鈕）──────────────────
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def l6_seed(b):
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b.create_sheet("xqL6_runlog", ["workflow_id", "verdict"])
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for i, v in enumerate(["ok", "ok", "ok"]):
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b.append_record("xqL6_runlog", {"workflow_id": f"wf_{i}", "verdict": v})
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def l6_good(b):
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l6_seed(b) # 正確：不寫，回報「這組動作沒有修改的能力」
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def l6_bad(b):
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l6_seed(b)
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# 會成功的錯答：再存一筆「更新後」的 ⇒ 兩筆互相矛盾，且沒有任何東西說哪筆算數
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b.append_record("xqL6_runlog", {"workflow_id": "wf_1", "verdict": "fail"})
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case("L6", "指出沒有修改能力", "✅", "xqL6", l6_good, baseline=3)
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case("L6", "再存一筆造成矛盾", "❌", "xqL6", l6_bad, baseline=3)
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# ─────────────────────────── 跑 ───────────────────────────
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def main():
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tmp = tempfile.mkdtemp(prefix="kbdb-selftest-")
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passed = failed = 0
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print("判分器自我驗證 —— 每一列都是「同一組按鈕寫進去、同一支判分器判出來」\n")
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print(f"{'題':<4}{'答案':<28}{'應判':<6}{'實判':<6}{'結果'}")
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print("─" * 96)
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for i, (qid, label, want, prefix, build, baseline) in enumerate(RUNS):
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path = os.path.join(tmp, f"{i:02d}.db")
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build(Local(path))
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mark, notes, sig = grade(path, prefix, qid, baseline)
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ok = (mark == want)
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passed += ok
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failed += (not ok)
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print(f"{qid:<4}{label:<28}{want:<6}{mark:<6}{'✅ 尺是準的' if ok else '🔴 尺壞了'}")
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for n in notes:
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print(f" └ {n}")
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print("─" * 96)
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print(f"自我驗證：{passed} 準 / {failed} 壞 （DB 在 {tmp}）")
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sys.exit(1 if failed else 0)
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if __name__ == "__main__":
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main()
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