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A subquery project provide dictionary data & service for a substrate-based network

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What is SubQuery?

SubQuery powers the next generation of Polkadot dApps by allowing developers to extract, transform and query blockchain data in real time using GraphQL. In addition to this, SubQuery provides production quality hosting infrastructure to run these projects in.

What is SubQuery Dictionary

This special SubQuery Project provides a dictionary of data that pre-indexes events on chain to dramatically improve indexing the performance of your own SubQuery Project, sometimes up to 10x faster.

It scans over the network, and records the module and method for every event/extrinsic on each block. See the standard entities in the schema.graphql file within this project to see what fields are available.

If you want to create your SubQuery Dictionary to speed up indexing of your own Substrate chain, fork this project and let us know

Getting Started

1. Add the dictionary URL to your SubQuery project

network:
  endpoint: wss://polkadot.api.onfinality.io/public-ws
  dictionary: https://api.subquery.network/sq/subquery/polkadot-dictionary

Understanding the dictionary

Visit How the dictionary works for more information.

Adding your own dictionary

  1. Fork the repository.
  2. Add your own dictionary project to the root directory.
  3. To publish on our Managed Services, you can use the GitHub actions.
    1. Set a GitHub environment to DEPLOYMENT.
    2. Set a GitHub secret name to SUBQL_ACCES_TOKEN (Your token generated on https://project.subquery.network/ ) under the previously set environment.
    3. Create a project using the UI or the CLI.
    4. Navigate to GitHub actions in your forked repository and select the workflow Deploy to Hosted Service.
      1. Enter your dictionary directory name in the input field Dictionary directory name e.g.acala.
      2. For the input field Project Name enter the name of the project created earlier (If your project name uses spaces please replace them with - e.g. perfect - dictionary -> perfect---dictionary).

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