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

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

从 FastMCP.resource() 装饰器的参数验证开始，检查 Context-only 函数是如何分类的。然后跟踪 Resource.read() 和现有的资源实现，以一致地更新调用签名；当 Context-only 资源出现在资源列表中，并且可以在不破坏现有资源的情况下访问它们时，即视为完成。