doaj-discovery-mcp
Rank #69,447glama/ramiramirez-nl/doaj-discovery-mcp
Costless-first Model Context Protocol server for discovering DOAJ-indexed journals and articles.
doaj-discovery-mcp is a Model Context Protocol (MCP) server published by ramiramirez-nl. It ranks #69,447 of 137,492 servers tracked on MCP Toplist, and its repository has 1 GitHub stars. doaj-discovery-mcp is listed on Glama, and ships as a single rolling release with no explicit version metadata. It was first listed on Jul 29, 2026.
Ranks ahead of 68,045 of 137,492 servers on MCP Toplist.
Use doaj-discovery-mcp
No credentials or required configuration declared — add it to your MCP client and go.
This server doesn't publish a machine-readable install config — see the repository README for install instructions.
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[](https://mcptoplist.com/server/glama%2Framiramirez-nl%2Fdoaj-discovery-mcp)<a href="https://mcptoplist.com/server/glama%2Framiramirez-nl%2Fdoaj-discovery-mcp"><img src="https://mcptoplist.com/badge/glama%2Framiramirez-nl%2Fdoaj-discovery-mcp.svg" alt="MCP Toplist: #69,447 of 137,492" /></a>Variants: append ?metric=score or ?metric=stars to the image URL.
Listed on 1 registry
ramiramirez-nl
Not versioned
This server is published through a registry that does not expose explicit version metadata, and no GitHub release tags were found on the linked repository. The listing tracks a single rolling release.
Frequently asked questions
- Who maintains doaj-discovery-mcp?
- doaj-discovery-mcp is maintained by ramiramirez-nl, which publishes 1 MCP server (0 total versions) tracked on MCP Toplist.
- Is doaj-discovery-mcp listed on the Official MCP Registry?
- doaj-discovery-mcp is not on the Official MCP Registry. It is listed on Glama.
- How many versions does doaj-discovery-mcp have?
- doaj-discovery-mcp ships as a single rolling release with no explicit version metadata.
- Where can I find the source code for doaj-discovery-mcp?
- The source code for doaj-discovery-mcp is hosted at github.com/ramiramirez-nl/doaj-discovery-mcp.
Do you run this server?
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