speclaw
Rank #9,548glama/esneiderbravo/speclaw
Local MCP server that gives AI agents a project constitution (LAWS.md), a code graph (Compass), and a spec-driven workflow — all 100% local with no LLM or cloud.
speclaw is a Model Context Protocol (MCP) server published by esneiderbravo. It ranks #9,548 of 132,345 servers tracked on MCP Toplist, and its repository has 1 GitHub stars. speclaw is listed on Glama, with 2 release tags tracked from its GitHub repository. It was first listed on Aug 2, 2026 and most recently updated on Sep 8, 2026.
Ranks ahead of 122,797 of 132,345 servers on MCP Toplist.
Use speclaw
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%2Fesneiderbravo%2Fspeclaw)<a href="https://mcptoplist.com/server/glama%2Fesneiderbravo%2Fspeclaw"><img src="https://mcptoplist.com/badge/glama%2Fesneiderbravo%2Fspeclaw.svg" alt="MCP Toplist: Top 10% of 132,345" /></a>Variants: append ?metric=score or ?metric=stars to the image URL.
Listed on 1 registry
esneiderbravo
GitHub releases (2)
No registry surfaces explicit version metadata for this server. The list below shows release tags from the linked GitHub repository.
| Version | Published |
|---|---|
| v1.0.1 | Sep 8, 2026 |
| v1.0.0 | Aug 25, 2026 |
Frequently asked questions
- Who maintains speclaw?
- speclaw is maintained by esneiderbravo, which publishes 1 MCP server (2 total versions) tracked on MCP Toplist.
- Is speclaw listed on the Official MCP Registry?
- speclaw is not on the Official MCP Registry. It is listed on Glama.
- How many versions does speclaw have?
- No registry surfaces explicit version metadata for speclaw; MCP Toplist tracks 2 release tags from its GitHub repository, most recently published on Sep 8, 2026.
- Where can I find the source code for speclaw?
- The source code for speclaw is hosted at github.com/esneiderbravo/speclaw.
Do you run this server?
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