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ThearchyHelios
9d375f8f1c
「测试 1」按校验提示改好之后仍然跑不完：while 循环配了 1005 轮，每轮是"循环节点 + 循环体"两步，而整次运行的步数上限（max_graph_steps）是 200——跑到 100 轮上下 LangGraph 抛 GraphRecursionError，报错是一段英文，叫人去调一个界面上碰不到的 recursion_limit。 图没写错，是平台自己的两个上限在打架：校验通过，跑到一半才死。 全局上限防的是没人把关的环；loop 节点的 max_iterations 是作者明说的轮数上限。现在循环 按自己的轮数另记一份预算（schema.loop_steps），加在全局上限之上，子图同理。取上界： 一个节点最多跑 Π(包住它的各层循环 max_iterations + 1) 次，嵌套相乘；done 出口之后的 节点只跑一次，不按轮数记账，否则预算虚高、等于没有上限。 没有 loop 节点把关的环（分支连回上游）照旧由全局上限拦住，报错改成中文、说清预算怎么 来的、该怎么改；子图里撞上限报子图自己的名字和步数——以前那段英文会被外层包进 NodeError 原样带出来。 实测：「测试 1」照提示改完，在真沙箱里 1000 轮跑满，25 秒，结果 501500（= 2+3+…+1001）。 测试：test_step_budget 8 条——预算的上界、嵌套相乘、done 之后不记账；全局上限压到 20 步 时循环照样跑满 30 轮；空转的环中文报错；子图里的循环有自己的预算、空转的子图报它自己的 名字。把 runner / 子图的预算换成 0，对应用例确实失败。README 补了常见问题，测试数更新为 470。
475 lines
23 KiB
Python
475 lines
23 KiB
Python
from __future__ import annotations
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import asyncio
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import json
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import time
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from typing import Any
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from langchain_core.messages import AIMessage, HumanMessage, SystemMessage, ToolMessage
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from langgraph.errors import GraphRecursionError
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from langgraph.func import task
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from app.core.config import settings
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from app.core.events import EventType
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from app.db.base import SessionLocal
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from app.db.models import Workflow
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from app.engine.context import NodeContext, NodeError
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from app.engine.state import GraphState, message_text, template_context
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from app.providers import catalog
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from app.providers.factory import ModelSpec, bind_tools_safely, get_chat_model
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from app.tools.registry import (
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ToolArgsError,
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ToolContext,
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args_model_of,
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build_tools,
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call_is_dangerous,
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prepare_args,
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)
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_MAX_DEPTH = 3
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# --------------------------------------------------------------------------
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# Supervisor：一个调度者 + 若干专家
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# --------------------------------------------------------------------------
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async def run_supervisor(state: GraphState, ctx: NodeContext) -> dict[str, Any]:
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"""多 agent 协作。
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调度者每轮只做一件事：看当前进展，决定交给哪个专家、并给它一句具体指令，
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或者宣布收工。专家各自带自己的 system prompt 和工具集。
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把整个协作塞进一个节点，是为了让画布保持可读 —— 展开细节看运行时间线就够了。
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"""
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agents = ctx.cfg("agents", []) or []
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if not agents:
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raise NodeError(ctx.node.id, "supervisor 节点至少要配一个 agent")
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names = [a.get("name") or f"agent{i}" for i, a in enumerate(agents)]
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agent_map = {n: a for n, a in zip(names, agents)}
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max_rounds = min(int(ctx.cfg("max_rounds", 6) or 6), settings.max_agent_steps)
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goal = ctx.render_str(ctx.cfg("goal", "{{ last_message }}"), state)
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if not goal.strip():
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goal = json.dumps(template_context(state).get("input"), ensure_ascii=False)
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async with SessionLocal() as session:
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supervisor_model, sup_model_id = await get_chat_model(
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session,
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ModelSpec(
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provider=ctx.cfg("provider"),
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model=ctx.cfg("model"),
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max_tokens=int(ctx.cfg("max_tokens", 2048) or 2048),
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),
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)
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roster = "\n".join(
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f"- {n}：{agent_map[n].get('description') or agent_map[n].get('system', '')[:100]}"
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for n in names
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)
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# 一轮可以派给多个专家。协议用 assignments 数组而不是单个 next：
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# 后者把"一次只能派一个"编进了协议本身，模型再想并行也表达不出来。
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#
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# 但并行只在任务**互不依赖**时才成立——并发的几个专家看到的是同一份进展
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# 快照，谁也看不见谁。派错了不会报错，只会让两个人基于同样的旧信息
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# 重复劳动，而这件事在结果里很难看出来。所以判断责任明确压在调度者身上，
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# 提示词里把反例写出来。
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route_schema = {
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"title": "route", # langchain 靠 title 识别 JSON Schema dict
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"type": "object",
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"properties": {
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"assignments": {
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"type": "array",
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"description": "这一轮要派出去的任务。互不依赖时可以给多个，它们会同时执行",
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"items": {
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"type": "object",
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"properties": {
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"agent": {"type": "string", "enum": names},
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"instruction": {"type": "string"},
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},
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"required": ["agent", "instruction"],
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},
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},
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"done": {"type": "boolean", "description": "目标已达成，结束协作"},
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"reason": {"type": "string"},
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},
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"required": ["assignments"],
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}
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#: 一轮最多同时派几个。再多的话调度者多半只是在把任务拆碎，
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#: 而每多一路就多一份"基于陈旧进展"的风险
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max_parallel = max(1, min(int(ctx.cfg("max_parallel", 3) or 3), len(names)))
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#: dangerous：需要人工确认的调用不执行（成员停不下来等人）；never：全部放行
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approval = ctx.cfg("approval", "dangerous")
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transcript: list[dict[str, Any]] = []
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usage_total = {"input_tokens": 0, "output_tokens": 0, "cost_usd": 0.0, "calls": 0}
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def _accumulate(msg: Any, model_id: str) -> None:
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meta = getattr(msg, "usage_metadata", None) or {}
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i, o = int(meta.get("input_tokens") or 0), int(meta.get("output_tokens") or 0)
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usage_total["input_tokens"] += i
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usage_total["output_tokens"] += o
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usage_total["calls"] += 1
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usage_total["cost_usd"] = round(
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usage_total["cost_usd"] + catalog.estimate_cost(model_id, i, o), 6
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)
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@task
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async def route(round_no: int, progress: str) -> dict[str, Any]:
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prompt = (
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f"你是团队调度者。目标：\n{goal}\n\n"
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f"可用成员：\n{roster}\n\n"
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f"当前进展：\n{progress or '(还没有任何进展)'}\n\n"
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"决定这一轮把什么任务交给谁，每个任务给一句明确的指令。\n\n"
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f"**互不依赖的任务可以放在同一轮**（最多 {max_parallel} 个），它们会同时执行，"
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"总耗时按最慢的那个算。判断依据只有一条：**后一个任务需不需要看到前一个的结果**。\n"
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" 可以同时派：「查 A 产品的资料」和「查 B 产品的资料」——两边互不相干\n"
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" 不能同时派：「查资料」和「根据资料写报告」——后者要等前者\n"
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" 不能同时派：「写初稿」和「校对初稿」——同上\n\n"
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"同时派出去的成员看到的是**同一份进展快照**，谁也看不见谁这一轮干了什么。"
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"拿不准是否独立时就分两轮派，代价只是慢一点；派错了则是两个人基于同样的"
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"旧信息重复劳动，而这在结果里很难看出来。\n\n"
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"目标已经达成时，done 填 true、assignments 给空数组。"
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)
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try:
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value = await supervisor_model.with_structured_output(route_schema).ainvoke(prompt)
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if not isinstance(value, dict):
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value = json.loads(message_text(value))
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except Exception: # noqa: BLE001
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# 不支持结构化输出的模型退回文本：从回复里认出一个成员名，按单人派
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raw = message_text(await supervisor_model.ainvoke(prompt))
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picked = next((n for n in names if n in raw), None)
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value = ({"assignments": [{"agent": picked, "instruction": raw[:500]}]}
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if picked else {"assignments": [], "done": True, "reason": raw[:200]})
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return value
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@task
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async def work(round_no: int, name: str, instruction: str, progress: str) -> dict[str, Any]:
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cfg = agent_map[name]
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async with SessionLocal() as session:
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model, model_id = await get_chat_model(
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session,
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ModelSpec(
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provider=cfg.get("provider") or ctx.cfg("provider"),
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model=cfg.get("model") or ctx.cfg("model"),
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max_tokens=int(cfg.get("max_tokens") or 4096),
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),
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)
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tool_ctx = ToolContext(
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run_id=ctx.run.run_id,
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node_id=f"{ctx.node.id}:{name}",
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sandbox_session=ctx.run.thread_id,
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memory_scope=ctx.run.memory_scope,
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collection=ctx.run.collection,
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)
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tools = await build_tools(cfg.get("tools", []) or [], tool_ctx, session=session)
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bound = bind_tools_safely(model, tools, parallel=False)
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tool_map = {t.name: t for t in tools}
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messages: list[Any] = [
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SystemMessage(content=cfg.get("system") or f"你是 {name}。"),
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HumanMessage(content=f"团队目标：{goal}\n\n已有进展：\n{progress or '(无)'}\n\n你的任务：{instruction}"),
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]
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calls: list[dict[str, Any]] = []
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# 和 agent 节点同一个硬顶：成员上配多少步都过不去 max_agent_steps
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max_steps = min(int(cfg.get("max_steps") or 4), settings.max_agent_steps)
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for _ in range(max_steps):
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reply = await bound.ainvoke(messages)
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_accumulate(reply, model_id)
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messages.append(reply)
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tool_calls = getattr(reply, "tool_calls", None) or []
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if not tool_calls:
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return {"text": message_text(reply), "tool_calls": calls}
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for call in tool_calls:
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tname = call.get("name", "")
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targs = call.get("args", {}) or {}
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cid = call.get("id") or f"{name}-{tname}"
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if tname not in tool_map:
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ctx.emit(EventType.TOOL_START, tool=tname, args=targs, agent=name, call_id=cid)
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content = f"错误：没有名为 {tname} 的工具"
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else:
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schema = args_model_of(tool_map[tname])
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fix_note = None
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args_error = None
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if schema is not None:
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try:
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targs, fix_note = prepare_args(schema, targs)
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except ToolArgsError as e:
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args_error = str(e)
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# 纠正后的参数才是真正要执行的那份，tool.start 要发它
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ctx.emit(EventType.TOOL_START, tool=tname, args=targs, agent=name, call_id=cid)
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if fix_note:
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ctx.emit(EventType.LOG, level="warn",
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message=f"工具 {tname}：{fix_note}", code="tool_args_fixed")
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if args_error:
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# 参数就不对，没必要真调一次。把"它接受什么"喂回去让它改
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content = args_error
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elif approval != "never" and call_is_dangerous(tool_map[tname], tname, targs):
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# 专家们并行跑在同一个节点里，停不下来等人：审批恢复时节点整个
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# 重放，几个人的 interrupt 谁先谁后对不上号。以前的做法是不审，
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# shell_exec / file_write / 可写库上的 DELETE 照跑不误——
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# agent 节点守着的那道门，在这里是敞开的。现在是不跑，并说清原因
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content = (
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f"没有执行：{tname} 这次调用需要人工确认，而协作团队里的成员"
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"不能停下来等人。换一种不需要它的做法；实在需要，交给团队外"
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"的 agent 节点去做（那里可以逐次审批）。"
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)
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ctx.emit(EventType.LOG, level="warn", code="tool_needs_approval",
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message=f"{name} 想调用 {tname}，需要人工确认，协作节点里不执行")
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else:
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try:
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raw = await tool_map[tname].ainvoke(targs)
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content = raw if isinstance(raw, str) else json.dumps(raw, ensure_ascii=False, default=str)
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except Exception as e: # noqa: BLE001
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content = f"工具失败：{type(e).__name__}: {e}"
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ctx.emit(EventType.TOOL_END, tool=tname, agent=name, call_id=cid, preview=content[:1500])
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calls.append({"tool": tname, "args": targs, "result": content[:2000]})
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messages.append(ToolMessage(content=content, tool_call_id=cid))
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# 步数用完了：最后一步一定是要了工具的，模型从没拿到过"说结论"的那一轮。
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# messages[-1] 是一条 ToolMessage——以前它就被当成这个成员的产出交给调度者，
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# 工具的原始返回冒充了结论。agent 节点修过同一个问题，这里照做：不给工具
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# （用没绑工具的 model，结构上就发不出调用）补一轮收尾。
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messages.append(HumanMessage(content=(
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f"步数预算用完了（{max_steps} 步），现在起不能再调用任何工具。基于已经查到的"
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"信息给出你这部分的结论，说清哪些没查到、结论因此有什么局限。不要编造数据。"
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)))
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try:
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settled = await model.ainvoke(messages)
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_accumulate(settled, model_id)
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text = message_text(settled).strip()
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except Exception as e: # noqa: BLE001 - 收尾失败不能把已有的过程一起赔进去
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ctx.emit(EventType.LOG, level="warn", code="settle_failed",
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message=f"{name} 的收尾轮没跑成：{type(e).__name__}: {e}")
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text = next((t for m in reversed(messages)
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if isinstance(m, AIMessage) and (t := message_text(m).strip())), "")
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ctx.emit(EventType.LOG, level="warn", code="step_limit_settled",
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message=f"{name} 用满了 {max_steps} 步，结论基于已经查到的部分")
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return {"text": text or f"（{name} 用满了 {max_steps} 步，没有给出结论）", "tool_calls": calls}
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progress_lines: list[str] = []
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final_text = ""
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rounds_meta: list[dict[str, Any]] = []
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for round_no in range(max_rounds):
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progress = "\n\n".join(progress_lines)
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decision = await route(round_no, progress)
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_accumulate(AIMessage(content=""), sup_model_id) # route 用了一次调用
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reason = str(decision.get("reason", ""))
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# 只留认得出的成员，并按上限截断。模型偶尔会重复派同一个人或者编一个
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# 不存在的名字——这两种在并发里都是纯浪费
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batch: list[tuple[str, str]] = []
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seen: set[str] = set()
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for item in decision.get("assignments") or []:
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name = str((item or {}).get("agent") or "")
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if name in agent_map and name not in seen:
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seen.add(name)
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batch.append((name, str(item.get("instruction") or goal)))
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if len(batch) >= max_parallel:
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break
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if decision.get("done") or not batch:
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# agents / done 是结构化字段，界面照着它渲染。以前只有一句中文
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# 消息串，解码层得拿正则去拆「调度 → X（理由）」——文案一改就散架
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ctx.emit(EventType.LOG, level="info", round=round_no,
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agents=[], done=True,
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message="调度 → 结束协作" + (f"（{reason}）" if reason else ""))
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final_text = progress_lines[-1].split("】", 1)[-1] if progress_lines else ""
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break
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who = "、".join(n for n, _ in batch)
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ctx.emit(
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EventType.LOG, level="info", round=round_no, parallel=len(batch),
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agents=[n for n, _ in batch], done=False, reason=reason,
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message=(f"调度 → {who}" + (f"（{reason}）" if reason else "")
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+ (f" · {len(batch)} 人同时进行" if len(batch) > 1 else "")),
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)
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for name, instruction in batch:
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ctx.emit(EventType.AGENT_STEP_START, agent=name, instruction=instruction[:300],
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round=round_no, parallel=len(batch))
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# 并发执行。它们拿到的是**同一份** progress 快照——这正是"任务必须
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# 互不依赖"的技术含义，也是上面提示词里那几条反例的由来。
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#
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# 每个人单独计时，而不是事后拿整轮墙钟去摊：界面要显示"并行省了多少
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# 时间"，那个数必须是量出来的。按 N×墙钟 推算等于假设每个人都跑满了
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# 最慢那条，而实际上快的那个可能只花了三分之一
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async def _timed(name: str, instruction: str) -> tuple[str, Any, int]:
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t0 = time.perf_counter()
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try:
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out: Any = await work(round_no, name, instruction, progress)
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except BaseException as exc: # noqa: BLE001 - 交给下面统一处置
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out = exc
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return name, out, int((time.perf_counter() - t0) * 1000)
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started = time.perf_counter()
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results = await asyncio.gather(*(_timed(n, i) for n, i in batch))
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wall_ms = int((time.perf_counter() - started) * 1000)
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sum_ms = 0
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for (name, instruction), (_, outcome, each_ms) in zip(batch, results):
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if isinstance(outcome, BaseException):
|
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# 一个专家挂了不该拖垮整轮：把失败当成它的产出交回调度者，
|
||
# 让它决定改派还是收工
|
||
text = f"（{name} 执行失败：{type(outcome).__name__}: {outcome}）"
|
||
outcome = {"text": text, "tool_calls": []}
|
||
ctx.emit(EventType.LOG, level="warn", round=round_no,
|
||
message=f"{name} 这一轮失败了：{text}")
|
||
sum_ms += each_ms
|
||
ctx.emit(EventType.AGENT_STEP_END, agent=name, duration_ms=each_ms,
|
||
round=round_no, parallel=len(batch), preview=outcome["text"][:500])
|
||
progress_lines.append(f"【{name}】{outcome['text']}")
|
||
transcript.append(
|
||
{"round": round_no, "agent": name, "instruction": instruction, **outcome,
|
||
"duration_ms": each_ms}
|
||
)
|
||
final_text = outcome["text"]
|
||
|
||
rounds_meta.append({
|
||
"round": round_no, "agents": [n for n, _ in batch],
|
||
"parallel": len(batch), "wall_ms": wall_ms, "sum_ms": sum_ms,
|
||
})
|
||
|
||
usage_total["total_tokens"] = usage_total["input_tokens"] + usage_total["output_tokens"]
|
||
# 并行省下的时间：各轮"串行本该花的"减去"实际花的"。只有并发过才有差值
|
||
saved_ms = sum(r["sum_ms"] - r["wall_ms"] for r in rounds_meta)
|
||
result = {
|
||
"text": final_text,
|
||
"rounds": len(rounds_meta),
|
||
"steps": len(transcript),
|
||
"transcript": transcript,
|
||
"agents": names,
|
||
"schedule": rounds_meta,
|
||
"parallel_saved_ms": saved_ms,
|
||
}
|
||
updates: dict[str, Any] = {
|
||
"nodes": {ctx.node.id: result},
|
||
"usage": usage_total,
|
||
}
|
||
if ctx.cfg("emit_message", True):
|
||
updates["messages"] = [AIMessage(content=final_text or "(空)")]
|
||
var_name = ctx.cfg("assign_to", "")
|
||
if var_name:
|
||
updates["vars"] = {var_name: final_text}
|
||
return updates
|
||
|
||
|
||
# --------------------------------------------------------------------------
|
||
# 子图：把另一张工作流当成一个节点
|
||
# --------------------------------------------------------------------------
|
||
|
||
|
||
async def run_subgraph(state: GraphState, ctx: NodeContext) -> dict[str, Any]:
|
||
"""嵌套执行另一个工作流。
|
||
|
||
子图共享同一个 run 的事件流（节点 id 会加前缀），所以在时间线上能看到它内部
|
||
每一步；但它有独立的状态，不会污染父图的变量池。
|
||
"""
|
||
from app.engine.compiler import compile_graph # 延迟导入，避免循环依赖
|
||
from app.engine.schema import GraphSpec, loop_steps
|
||
|
||
if ctx.run.depth >= _MAX_DEPTH:
|
||
raise NodeError(ctx.node.id, f"子图嵌套超过 {_MAX_DEPTH} 层，已阻止")
|
||
|
||
workflow_id = ctx.cfg("workflow_id")
|
||
if not workflow_id:
|
||
raise NodeError(ctx.node.id, "子图节点没有选择工作流")
|
||
|
||
# 版本钉死：这就是"方法卡"的机制核心。钉了 workflow_version 就永远
|
||
# 执行那个不可变快照，上游改了方法卡也不会让这里的口径悄悄漂移；
|
||
# 没钉则跟随最新（画布试跑方便，但治理 lint 会在受管模板里拦下它）。
|
||
pinned = ctx.cfg("workflow_version")
|
||
async with SessionLocal() as session:
|
||
workflow = await session.get(Workflow, workflow_id)
|
||
if not workflow:
|
||
raise NodeError(ctx.node.id, f"找不到工作流 {workflow_id}")
|
||
if pinned:
|
||
from sqlalchemy import select
|
||
|
||
from app.db.models import WorkflowVersion
|
||
|
||
snapshot = (
|
||
await session.execute(
|
||
select(WorkflowVersion).where(
|
||
WorkflowVersion.workflow_id == workflow_id,
|
||
WorkflowVersion.version == int(pinned),
|
||
)
|
||
)
|
||
).scalar_one_or_none()
|
||
if not snapshot:
|
||
raise NodeError(
|
||
ctx.node.id, f"工作流「{workflow.name}」没有 v{pinned} 这个版本"
|
||
)
|
||
sub_graph = snapshot.graph
|
||
used_version = int(pinned)
|
||
else:
|
||
sub_graph = workflow.graph
|
||
used_version = workflow.version
|
||
|
||
sub_input = ctx.render(ctx.cfg("input", {}) or {}, state)
|
||
if not isinstance(sub_input, dict):
|
||
sub_input = {"input": sub_input}
|
||
|
||
sub_spec = GraphSpec.model_validate(sub_graph)
|
||
sub_run = type(ctx.run)(
|
||
run_id=ctx.run.run_id,
|
||
thread_id=f"{ctx.run.thread_id}:{ctx.node.id}",
|
||
spec=sub_spec,
|
||
workflow_id=workflow.id,
|
||
depth=ctx.run.depth + 1,
|
||
memory_scope=ctx.run.memory_scope,
|
||
collection=ctx.run.collection,
|
||
extra={**ctx.run.extra, "node_prefix": f"{ctx.node.id}/"},
|
||
)
|
||
|
||
ctx.emit(EventType.LOG, level="info",
|
||
message=f"进入子图「{workflow.name}」（{len(sub_spec.nodes)} 个节点）")
|
||
|
||
compiled = compile_graph(sub_spec, sub_run)
|
||
# 子图不带 checkpointer：它的断点由父图的 checkpoint 统一承载
|
||
app = compiled.compile()
|
||
limit = settings.max_graph_steps + loop_steps(sub_spec)
|
||
try:
|
||
result_state = await app.ainvoke(
|
||
{
|
||
"input": sub_input,
|
||
"vars": sub_input,
|
||
"nodes": {},
|
||
"output": {},
|
||
"messages": [],
|
||
"usage": {},
|
||
"trail": [],
|
||
"loops": {},
|
||
},
|
||
{"recursion_limit": limit},
|
||
)
|
||
except GraphRecursionError as e:
|
||
# 不接住的话，外层包装成 NodeError 时带的是 LangGraph 的英文原文
|
||
raise NodeError(
|
||
ctx.node.id,
|
||
f"子图「{workflow.name}」走满了 {limit} 步还没跑完，已中止。多半是子图里有个环"
|
||
"在空转：分支连回了上游、却没有 loop 节点给它定轮数上限",
|
||
) from e
|
||
|
||
output = result_state.get("output") or {}
|
||
result = {
|
||
"output": output,
|
||
"workflow": workflow.name,
|
||
# 溯源信息：这次到底跑的是哪个版本、是不是钉死的
|
||
"workflow_version": used_version,
|
||
"pinned": bool(pinned),
|
||
"text": json.dumps(output, ensure_ascii=False) if output else "",
|
||
}
|
||
ctx.emit(EventType.LOG, level="info", message=f"子图「{workflow.name}」完成")
|
||
|
||
updates: dict[str, Any] = {
|
||
"nodes": {ctx.node.id: result},
|
||
"usage": result_state.get("usage") or {},
|
||
}
|
||
var_name = ctx.cfg("assign_to", "")
|
||
if var_name:
|
||
updates["vars"] = {var_name: output}
|
||
return updates
|