How to filter generated Swagger JSON (yaml)

问题

I have over 5k lines long swagger.json file describing hundreds of paths and objects. I want to generate a TypeScript client (using swagger-codegen) using only a part of the endpoints. I don't want the generated TypeScript application to contain classes or interfaces connected with unused part of the swagger.json

How to filter out only a part of the Swagger documentation, describing the specified group of paths (e.g. all paths starting with /api/*)? Especially I want the filtered JSON to not contain definitions for unused data structures.

回答1:

I think you can do it, using task automation (grunt, gulp, shell, whatever). Basically it could be a 3 steps task:

1. get the swagger.json (or call the swagger code-gen to get the json, with something like java -jar swagger-codegen-cli-x.x.x.jar generate -i <URL> -l swagger -o GeneratedCodeSwagger )
2. remove the definitions/paths that you want to exclude and create a modified swagger.json
3. call the code-gen passing the modified json with java -jar swagger-codegen-cli-x.x.x.jar generate -i GeneratedCodeSwagger\swagger.json -l typescript-angular

回答2:

Finally, we created swagger-json-filter – a command-line tool allowing filtering a Swagger documentation. It can be easily used along other commands in bash:

cat input.json | swagger-json-filter --include-paths="^\/api\/.*" > output.json

The tool is performing a logic needed to filter out undesired definitions (nested also) from the output.

来源：https://stackoverflow.com/questions/39788099/how-to-filter-generated-swagger-json-yaml