Interrupt(). How to interrupt for the second time

[Saisiva123]

Checked other resources

• This is a bug, not a usage question. For questions, please use GitHub Discussions.
• I added a clear and detailed title that summarizes the issue.
• I read what a minimal reproducible example is (https://stackoverflow.com/help/minimal-reproducible-example).
• I included a self-contained, minimal example that demonstrates the issue INCLUDING all the relevant imports. The code run AS IS to reproduce the issue.

Example Code

def ask_user_node(state: LookupState) -> Command[Literal['lookup_node']]:
        agent_request = state['userAgentInteractionInfo']['agentRequest']

        if 'userResponse' in state['userAgentInteractionInfo']:
            del state['userAgentInteractionInfo']['userResponse']

        user_response = interrupt({
            'id': str(uuid.uuid4()),
            'request': agent_request
        })

        if user_response:
            return Command(goto='lookup_node',
                           resume={'userAgentInteractionInfo': {'agentRequest': agent_request}},
                           update={'messages': [HumanMessage(content=user_response, name="User_Response")], 'userAgentInteractionInfo': {'agentRequest': '', 'userResponse': ''}})

Error Message and Stack Trace (if applicable)

No response

Description

I know the default behavior of interrupt will invoke GraphInterrupt for the first time. When the same node has the interrupt its not halting the execution instead its taking the same old cached value, so is there any extra argument that I can pass so that the interrupt will halt the execution for the second time and wait for the user input

System Info

python -m langchain_core.sys_info

[Saisiva123]

There are many developers who are facing this issue, it would be nicer for the langgraph team to consider this issue as priority.

[vbarda]

This is a duplicate of #2935