第四单元——第二课 LangGraph 的“条件分支”

这一课学 LangGraph 的“条件分支”检索到资料就回答没检索到就走兜底分支。关键是add_conditional_edges()——它根据当前状态决定下一步运行哪个节点。官方说明在langchain-study文件夹新建unit4_lesson2_conditional_graph.pyfrom typing import Literal, NotRequired, TypedDict from langgraph.graph import START, END, StateGraph class BookstoreState(TypedDict): question: str context: NotRequired[str] answer: NotRequired[str] def retrieve(state: BookstoreState) - dict[str, str]: print(执行retrieve) if 会员 in state[question]: return {context: 会员购买图书可以享受八折优惠。} if 退货 in state[question]: return {context: 购买后七天内可凭购物小票申请退货。} return {context: } def choose_route( state: BookstoreState, ) - Literal[answer_found, answer_missing]: return answer_found if state[context] else answer_missing def answer_found(state: BookstoreState) - dict[str, str]: print(执行answer_found) return {answer: state[context]} def answer_missing(state: BookstoreState) - dict[str, str]: print(执行answer_missing) return {answer: 资料中没有相关信息。} builder StateGraph(BookstoreState) builder.add_node(retrieve, retrieve) builder.add_node(answer_found, answer_found) builder.add_node(answer_missing, answer_missing) builder.add_edge(START, retrieve) builder.add_conditional_edges(retrieve, choose_route) builder.add_edge(answer_found, END) builder.add_edge(answer_missing, END) graph builder.compile() for question in [会员买书有什么优惠, 书店几点关门]: print(f\n问题{question}) result graph.invoke({question: question}) print(回答, result[answer]) print(\n图的 Mermaid 表示) print(graph.get_graph().draw_mermaid())运行python unit4_lesson2_conditional_graph.py你应该看到第一个问题执行answer_found第二个执行answer_missing。注意这里不要再加retrieve → answer_found的普通边否则它可能和条件分支一起执行。这课的核心问题choose_route是在retrieve之前还是之后执行跑完看输出我们再继续。