Swagger能否生成与C#输入模型一致的泛型输出模型？

Absolutely! You can get Swagger Codegen to generate a generic TypeScript interface that matches your C# ApiResult<T> structure. The issue you're seeing happens because Swagger (by default) expands generic type instances into concrete, non-generic names like ApiResultString, and older Codegen configurations don't preserve generic definitions. Here's how to fix it step by step:

Step 1: Configure Swagger to Preserve Generic Type Definitions #

First, make sure your C# Swagger setup (using Swashbuckle) outputs the generic type structure instead of expanding it into concrete names.

In your .NET project's startup configuration (either Startup.cs for .NET 5 and below, or Program.cs for .NET 6+):

services.AddSwaggerGen(c =>
{
    // Include XML comments to help Swagger understand your types (optional but recommended)
    var xmlFile = $"{Assembly.GetExecutingAssembly().GetName().Name}.xml";
    var xmlPath = Path.Combine(AppContext.BaseDirectory, xmlFile);
    c.IncludeXmlComments(xmlPath);

    // Customize schema IDs to retain generic type syntax (e.g., ApiResult<T> instead of ApiResultString)
    c.CustomSchemaIds(type => 
    {
        if (type.IsGenericType)
        {
            // Extract the base generic name (strip the `1` suffix) and list generic arguments
            var genericBaseName = type.Name.Split('`')[0];
            var genericArgs = string.Join(",", type.GetGenericArguments().Select(t => t.Name));
            return $"{genericBaseName}<{genericArgs}>";
        }
        return type.Name;
    });
});

After this change, your Swagger schema will show ApiResult<string> instead of ApiResultString—this is key for Codegen to recognize the generic structure.

Step 2: Configure Swagger Codegen to Generate Generic Interfaces #

Next, tell Swagger Codegen to generate generic TypeScript types instead of concrete ones. Use the supportsGenericInterfaces=true additional property when running Codegen.

Example Command Line Usage: #

swagger-codegen generate -i path/to/your/swagger.json -l typescript-angular -o ./generated-types --additional-properties=supportsGenericInterfaces=true

Example Maven Plugin Configuration: #

If you're using the Maven plugin, add the property to your pom.xml:

<build>
    <plugins>
        <plugin>
            <groupId>io.swagger.codegen.v3</groupId>
            <artifactId>swagger-codegen-maven-plugin</artifactId>
            <version>3.0.34</version>
            <executions>
                <execution>
                    <goals>
                        <goal>generate</goal>
                    </goals>
                    <configuration>
                        <inputSpec>path/to/your/swagger.json</inputSpec>
                        <language>typescript-angular</language>
                        <output>./generated-types</output>
                        <additionalProperties>supportsGenericInterfaces=true</additionalProperties>
                    </configuration>
                </execution>
            </executions>
        </plugin>
    </plugins>
</build>

Step 3: Verify the Generated TypeScript Code #

Once you run Codegen with these settings, you'll get a generic interface that matches your original C# model:

'use strict';
export interface ApiResult<T> {
    result?: T;
    success?: boolean;
}

You can now use this exactly like your C# generic type—e.g., ApiResult<string>, ApiResult<YourCustomModel>, etc.

Key Notes: #

• Use Updated Versions: Ensure you're using a recent version of Swashbuckle.AspNetCore (v5+ recommended) and Swagger Codegen (v3+), as older versions have limited generic support.
• Template Compatibility: The supportsGenericInterfaces property works with most TypeScript templates (like typescript-angular, typescript-node, typescript-fetch). Double-check your template's documentation if you run into issues.
• Manual Swagger JSON Fix (If Needed): If you can't adjust the .NET Swagger setup, you can manually edit your swagger.json to replace concrete generic instances with a generic schema definition before running Codegen.

内容的提问来源于stack exchange，提问作者MelnikovI