Multiple Openapi specs?

[jbnitorum]

Is there any clean way to combine multiple servers created from OpenAPI specs or rather use multiple OpenAPIs specs to create a single server? Perhaps I'm missing it somewhere. Maybe a different way to get there is one server using from_openapi and then if there was something like Tool.from_openapi_route() that could create an additional tool to add.

I'm sure there a host of people that will tell me to just make two separate servers and turn them both on. The problem is when it comes to prompts. If I have prompts on the server that ask for tool calls from multiple servers...

1. On what server should these prompts live? They uses tools on server A and server B... I have to choose one?
2. Seems hardly perfect that I could ensure someone is actually running server B if they run a prompt from server A. This feels like shipping something without a necessary dependency.

Obviously I could just manually create the tools within server A that would otherwise make up server B... but that really takes some of the magic out of this for me...

[jbnitorum]

Answering my own question. This feels a little silly but it works if it helps someone else.

import httpx
from fastmcp import FastMCP
import asyncio

API_KEY = "XXXXXX"
openapi_spec = httpx.get("https://example1.com/openapi.json", headers={'x-api-key': API_KEY}).json()
client = httpx.AsyncClient(base_url="https://example1.com", headers={'x-api-key': API_KEY})

API_KEY_2 = "XXXXXX"
openapi_spec2 = httpx.get("https://example2.com/openapi.json").json()
client2 = httpx.AsyncClient(base_url="https://example2.com", headers={'x-api-key': API_KEY_2})    

mcp = FastMCP.from_openapi(
    openapi_spec=openapi_spec,
    client=client,
    name="The Actual Server",
)

mcp2 = FastMCP.from_openapi(
    openapi_spec=openapi_spec2,
    client=client2,
    name="The server thats a temporary vehicle to leverage OpenAPI integration"
)

server2_tools = asyncio.run(mcp2.get_tools())

for key in server2_tools.keys():
    mcp.add_tool(server2_tools[key])

[duanckham]

Same question.