It is not possible to define a Resource that takes only the Context parameter as an argument.

FastMCP.resource() デコレーターのパラメーター検証から始め、Context-only 関数がどのように分類されるかを調べます。次に Resource.read() と既存のリソース実装を追跡し、呼び出しシグネチャーを一貫して更新します。Context-only リソースがリソース一覧に表示され、既存のリソースを壊すことなくアクセスできれば完了です。

Initial Checks

• I confirm that I'm using the latest version of MCP Python SDK
• I confirm that I searched for my issue in https://github.com/modelcontextprotocol/python-sdk/issues before opening this issue

Description

Problem

Currently, it is not possible to define a Resource that takes only the Context parameter as an argument. When attempting to create such a resource, the registration logic incorrectly classifies it as a template resource instead of a regular resource.

Expected Behavior

Resources with only a Context parameter should be registered as regular resources, not template resources.

Current Behavior

Resources with only a Context parameter are incorrectly registered as template resources, causing them to not appear in the resource list and fail when accessed.

Root Cause

The parameter validation logic in the FastMCP.resource() decorator doesn't properly distinguish between effective function parameters and Context parameters. It treats any function with parameters as a potential template resource, even when the only parameter is a Context.

Solution

1. Fix the parameter validation logic to exclude Context parameters when determining if a resource should be a template
2. Update the Resource.read() method signature to accept an optional context parameter
3. Update all resource implementations to support the new signature

This change maintains full backward compatibility while enabling context-only resources.

Example Code

@mcp.resource("resource://user_profile")
def get_user_profile(ctx: Context) -> str:
    # Something Code ...

Python & MCP Python SDK

Python 3.13.5
MCP Python SDK v1.15.0