chore(bug/qa): 归档测试清单、bug 报告与 QA 诊断脚本

- core/bug: 各轮电商项目测试清单 xlsx + 验证报告/套图提示词梳理 md
- core/qa: consumption_summary / diag_login / diag_sessions / probe_usage 诊断脚本

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>

core/qa/consumption_summary.py

@@ -0,0 +1,79 @@
+"""消费汇总:把 DB 里所有扣费/充值/退费流水按团队、按任务类型、按模型聚合。
+跑法:cd core/backend && DJANGO_SETTINGS_MODULE=airshelf.settings.development .venv/bin/python ../qa/consumption_summary.py
+"""
+import os, sys
+from collections import defaultdict
+from decimal import Decimal
+
+sys.path.insert(0, os.path.join(os.path.dirname(__file__), "..", "backend"))
+os.environ.setdefault("DJANGO_SETTINGS_MODULE", "airshelf.settings.development")
+import django  # noqa: E402
+
+django.setup()
+
+from django.db.models import Sum, Count  # noqa: E402
+from apps.billing.models import CreditLedger  # noqa: E402
+from apps.ai.models import AITask  # noqa: E402
+
+Z = Decimal("0")
+
+print("=" * 60)
+print("一、全平台流水汇总(内部额度,单位 CNY)")
+print("=" * 60)
+for row in (
+    CreditLedger.objects.values("ledger_type")
+    .annotate(total=Sum("amount"), n=Count("id"))
+    .order_by("ledger_type")
+):
+    print(f"  {row['ledger_type']:12} 笔数={row['n']:6}  合计={row['total'] or Z}")
+
+charge_total = CreditLedger.objects.filter(ledger_type="charge").aggregate(s=Sum("amount"))["s"] or Z
+recharge_total = CreditLedger.objects.filter(ledger_type="recharge").aggregate(s=Sum("amount"))["s"] or Z
+refund_total = CreditLedger.objects.filter(ledger_type="refund").aggregate(s=Sum("amount"))["s"] or Z
+print(f"\n  >>> 实际消费(charge)总额 = {charge_total}")
+print(f"  >>> 充值(recharge)总额   = {recharge_total}")
+print(f"  >>> 退费(refund)总额      = {refund_total}")
+
+print("\n" + "=" * 60)
+print("二、按团队拆分消费(charge)")
+print("=" * 60)
+for row in (
+    CreditLedger.objects.filter(ledger_type="charge")
+    .values("team__name", "team_id")
+    .annotate(total=Sum("amount"), n=Count("id"))
+    .order_by("-total")
+):
+    print(f"  team={row['team__name']!r:24} 任务数={row['n']:5}  消费={row['total']}")
+
+print("\n" + "=" * 60)
+print("三、按任务类型拆分消费(charge 关联的 AITask.task_type)")
+print("=" * 60)
+agg = defaultdict(lambda: [0, Z])  # task_type -> [count, sum]
+qs = (
+    CreditLedger.objects.filter(ledger_type="charge", task__isnull=False)
+    .select_related("task")
+    .values_list("task__task_type", "amount")
+)
+for tt, amt in qs:
+    agg[tt][0] += 1
+    agg[tt][1] += amt or Z
+for tt, (n, s) in sorted(agg.items(), key=lambda kv: -kv[1][1]):
+    print(f"  {tt:16} 次数={n:6}  消费={s}")
+
+print("\n" + "=" * 60)
+print("四、单价表(ModelConfig.unit_price = 平台每次固定收费,非三方实际成本)")
+print("=" * 60)
+from apps.ai.models import ModelConfig  # noqa: E402
+for mc in ModelConfig.objects.all().order_by("capability", "name"):
+    print(f"  {mc.capability:8} {mc.provider.name}:{mc.name:30} 单价={mc.unit_price}  status={mc.status}")
+
+print("\n" + "=" * 60)
+print("五、AITask 成本字段抽样(estimated_cost vs actual_cost)")
+print("=" * 60)
+print(f"  task 总数={AITask.objects.count()}  成功={AITask.objects.filter(status='succeeded').count()}  失败={AITask.objects.filter(status='failed').count()}")
+print(f"  actual_cost 总和={AITask.objects.aggregate(s=Sum('actual_cost'))['s'] or Z}")
+# 检查 response_payload 里有没有三方返回的 usage/token/费用
+sample = AITask.objects.filter(status="succeeded").exclude(response_payload={}).first()
+if sample:
+    keys = list(sample.response_payload.keys()) if isinstance(sample.response_payload, dict) else type(sample.response_payload)
+    print(f"  response_payload 样本 keys = {keys}")