Swagger UI 不会在控制器的 GET 操作中为我的复杂类型参数呈现主体参数字段

Sadly, you can't do what you want with Swagger. You can't send a request model in an HTTP GET method. You can however change the swagger UI to look like this:

but you won't be able to receive the model in your controller.

This is a known issue within the Swagger developers and it was discussed in 2016 and the final decision is that swagger won't support a request body in an HTTP GET method. Here is the link to the already closed issue.

You have three options here:

• Leave the method as it is, and test it in Postman, but not in Swagger.
• 请按照以下步骤实现上图，但请注意，它只会修复 UI 部分，并且null SearchModel当您按下Try it out!swagger时，您将始终在控制器中结束。
• 使它成为一种[HttpPost方法而不是[HttpGet].

如何使 swagger UI 显示带有请求正文的 GET 方法：

首先，创建一个Attribute类：

public class ModelInBodyAttribute : Attribute
{
    public ModelInBodyAttribute(string modelName, string description, bool isRequired)
    {
        this.ModelName = modelName;
        this.Description = description;
        this.IsRequired = IsRequired;
    }

    public string ModelName { get; set; }
    public bool IsRequired { get; set; }
    public string Description { get; set; }
}

然后你可以在控制器中装饰你的方法：

[ModelInBody(modelName: nameof(SearchModel), description: "My model description", isRequired: true)]
[HttpGet]
[Route("search")]
[ResponseType(typeof(List<SearchModel>))]
public IHttpActionResult Search(SearchModel searchOptions)
{
    //....

    return Ok(new List<SearchModel>());
}

之后创建IOperationFilter类（ModelInBodyOperationFilter）：

public class ModelInBodyOperationFilter : IOperationFilter
{
    public void Apply(Operation operation, SchemaRegistry schemaRegistry, ApiDescription apiDescription)
    {
        var attribute = apiDescription.GetControllerAndActionAttributes<ModelInBodyAttribute>().FirstOrDefault();
        if (attribute == null)
        {
            return;
        }

        operation.parameters.Clear();
        operation.parameters.Add(new Parameter
        {
            name = attribute.ModelName,
            description = attribute.Description,
            @in = "body",
            required = attribute.IsRequired,
            schema = new Schema { @ref = $"#/definitions/{attribute.ModelName}" }
        });
    }
}

最后，不要忘记注册IOperationFilterin SwaggerConfig：

c.OperationFilter<ModelInBodyOperationFilter>();

当您通过 swagger 发送请求时，您会注意到该Curl部分是绝对正确的，但仍然在您的控制器中没有任何内容。