Skip to main content
Glama
symlink_only_dir_entry.py800 B
#!/usr/bin/env python3 # Copyright (c) Meta Platforms, Inc. and affiliates. # # This source code is licensed under both the MIT license found in the # LICENSE-MIT file in the root directory of this source tree and the Apache # License, Version 2.0 found in the LICENSE-APACHE file in the root directory # of this source tree. # Simple python script that symlinks a provided path to the only file in a directory import os import sys def main(): d = sys.argv[1] linkname = sys.argv[2] entries = os.listdir(d) if len(entries) != 1: raise ValueError("expected exactly one entry in directory") dest = os.path.join(d, entries[0]) symlink = os.path.relpath(dest, start=os.path.dirname(linkname)) os.symlink(symlink, linkname) if __name__ == "__main__": main()

Latest Blog Posts

MCP directory API

We provide all the information about MCP servers via our MCP API.

curl -X GET 'https://glama.ai/api/mcp/v1/servers/systeminit/si'

If you have feedback or need assistance with the MCP directory API, please join our Discord server