swagger-api / swagger-api/swagger-codegen

how to generate only model using the online generator

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Description

Description

apologies - this is more of a question: with the swagger-codegen-cli, there's a way to generate only the models by providing the -D model {opts} parameter.
Is there an equivalent json field that i can provide when using the online generator to do the same?

Swagger-codegen version

V3

Swagger declaration file content or url

None

Command line used for generation

currently i use the following:
curl -X POST
https://generator3.swagger.io/api/generate
-H 'content-type: application/json'
-d '{
"specURL" : "https://s3-us-west-2.amazonaws.com/aoeanpss3.prod.aws.ean/ean-rapid-site/documentation/rapid_2.3/swagger.yaml",
"lang" : "typescript-angular",
"options": {
"additionalProperties": {
"modelPropertyNaming": "original"
}
},
"type" : "CLIENT",
"codegenVersion" : "V3"
}'

Steps to reproduce

Not a bug

Related issues/PRs

NA

Suggest a fix/enhancement

NA

Contributor guide

Open the contributing guide

First steps

  1. Read the whole issue, then the project's contributing guide.
  2. Comment on the issue to say you are picking it up — it saves two people doing the same work.
  3. Fork the repository and make your change on a branch.
  4. Open a pull request that references the issue number.

Research direction

Start with the POST /api/generate request shown in the issue and inspect the online generator's API documentation or request schema for model-only generation options. Confirm whether an equivalent field exists for the TypeScript Angular client, then document the request setting and an example; done means the documented request is verified against the V3 generator.

Written by the indexing model from the issue text.

Assessment

Tech stack
angular, typescript
Domain
api, tooling
Issue type
Documentation
Difficulty
2/5
Estimated time
1-3 hours
Activity status
Stale
Clarity
Needs clarification
Newbie friendliness
25/100

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