swagger-api / swagger-api/swagger-codegen
[csharp] Generate a List<Guid> Model property instead of List<Guid?>
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Description
Description
The following model generates a class with a UUIDs property, but the type of the property is List<Guid?>. How can I generate a List<Guid>?
Swagger-codegen version
2.2.3
Swagger declaration file content or url
SearchModel:
type: object
description: A request model.
required: [uuids]
properties:
uuids:
type: array
items:
type: string
format: uuid
description: The unique identifier.
Command line used for generation
java -Dapis -Dmodels -DsupportingFiles -DapiTests=false -DmodelDocs=true -DmodelTests=false 2>>%4 -jar %5\swagger-codegen-cli-2.2.3.jar generate -i "%%f" -l csharp -o %2 --config %3 -t %5\templates
Output
The above yaml generates a Model with the following UUIDs property.
/// <summary>
/// Gets or Sets Uuids
/// </summary>
[DataMember(Name="uuids", EmitDefaultValue=false)]
public List<Guid?> Uuids { get; set; }
Contributor guide
First steps
- Read the whole issue, then the project's contributing guide.
- Comment on the issue to say you are picking it up — it saves two people doing the same work.
- Fork the repository and make your change on a branch.
- Open a pull request that references the issue number.
Research direction
Reproduce the generated C# model from the supplied Swagger YAML and command, then trace the C# model generation entry point and its type-mapping or template handling for required UUID arrays. Done means the generated Uuids property is List rather than List<Guid?> while preserving the declared required-array behavior.
Written by the indexing model from the issue text.
Assessment
- Tech stack
- csharp
- Domain
- tooling
- Issue type
- Bug
- Difficulty
- 3/5
- Estimated time
- 1-2 days
- Activity status
- Stale
- Clarity
- Mostly clear
- Newbie friendliness
- 38/100