是否有一种方法显示所有枚举作为他们的字符串值在swagger而不是他们的int值?
我希望能够提交POST动作，并根据它们的字符串值放置枚举，而不必每次都查看enum。
我尝试了DescribeAllEnumsAsStrings，但服务器然后接收字符串而不是enum值，这不是我们要寻找的。
有人解决了吗?
编辑:
public class Letter
{
[Required]
public string Content {get; set;}
[Required]
[EnumDataType(typeof(Priority))]
public Priority Priority {get; set;}
}
public class LettersController : ApiController
{
[HttpPost]
public IHttpActionResult SendLetter(Letter letter)
{
// Validation not passing when using DescribeEnumsAsStrings
if (!ModelState.IsValid)
return BadRequest("Not valid")
..
}
// In the documentation for this request I want to see the string values of the enum before submitting: Low, Medium, High. Instead of 0, 1, 2
[HttpGet]
public IHttpActionResult GetByPriority (Priority priority)
{
}
}
public enum Priority
{
Low,
Medium,
High
}
我得到了它的工作。net 6 Web API使用以下代码从其他答案在这里:
1 -创建一个DocumentFilter
/// <summary>
/// Add enum value descriptions to Swagger
/// </summary>
public class SwaggerEnumDocumentFilter : IDocumentFilter
{
public void Apply(OpenApiDocument swaggerDoc, DocumentFilterContext context)
{
// add enum descriptions to result models
foreach (var property in swaggerDoc.Components.Schemas)
{
var propertyEnums = property.Value.Enum;
if (propertyEnums is { Count: > 0 })
{
property.Value.Description += DescribeEnum(propertyEnums, property.Key);
}
}
if (swaggerDoc.Paths.Count <= 0)
{
return;
}
// add enum descriptions to input parameters
foreach (var pathItem in swaggerDoc.Paths.Values)
{
DescribeEnumParameters(pathItem.Parameters);
var affectedOperations = new List<OperationType> { OperationType.Get, OperationType.Post, OperationType.Put, OperationType.Patch };
foreach (var operation in pathItem.Operations)
{
if (affectedOperations.Contains(operation.Key))
{
DescribeEnumParameters(operation.Value.Parameters);
}
}
}
}
private static void DescribeEnumParameters(IList<OpenApiParameter> parameters)
{
if (parameters == null) return;
foreach (var param in parameters)
{
if (param.Schema.Reference != null)
{
var enumType = GetEnumTypeByName(param.Schema.Reference.Id);
var names = Enum.GetNames(enumType).ToList();
param.Description += string.Join(", ", names.Select(name => $"{Convert.ToInt32(Enum.Parse(enumType, name))} - {name}").ToList());
}
}
}
private static Type GetEnumTypeByName(string enumTypeName)
{
if (string.IsNullOrEmpty(enumTypeName))
{
return null;
}
try
{
return AppDomain.CurrentDomain
.GetAssemblies()
.SelectMany(x => x.GetTypes())
.Single(x => x.FullName != null
&& x.Name == enumTypeName);
}
catch (InvalidOperationException e)
{
throw new Exception($"SwaggerDoc: Can not find a unique Enum for specified typeName '{enumTypeName}'. Please provide a more unique enum name.");
}
}
private static string DescribeEnum(IEnumerable<IOpenApiAny> enums, string propertyTypeName)
{
var enumType = GetEnumTypeByName(propertyTypeName);
if (enumType == null)
{
return null;
}
var parsedEnums = new List<OpenApiInteger>();
foreach (var @enum in enums)
{
if (@enum is OpenApiInteger enumInt)
{
parsedEnums.Add(enumInt);
}
}
return string.Join(", ", parsedEnums.Select(x => $"{x.Value} - {Enum.GetName(enumType, x.Value)}"));
}
}
2 -将其添加到Program.cs文件中
services.AddSwaggerGen(config =>
{
config.DocumentFilter<SwaggerEnumDocumentFilter>();
})
在这里搜索答案后，我发现了在模式中显示枚举的问题的部分解决方案[SomeEnumString = 0, AnotherEnumString = 1]，但与此相关的所有答案都只是部分正确的，正如@OhWelp在其中一个评论中提到的那样。
下面是。net core的完整解决方案(2、3，目前在6上工作):
public class EnumSchemaFilter : ISchemaFilter
{
public void Apply(OpenApiSchema model, SchemaFilterContext context)
{
if (context.Type.IsEnum)
{
model.Enum.Clear();
var names = Enum.GetNames(context.Type).ToList();
names.ForEach(name => model.Enum.Add(new OpenApiString($"{GetEnumIntegerValue(name, context)} = {name}")));
// the missing piece that will make sure that the new schema will not replace the mock value with a wrong value
// this is the default behavior - the first possible enum value as a default "example" value
model.Example = new OpenApiInteger(GetEnumIntegerValue(names.First(), context));
}
}
private int GetEnumIntegerValue(string name, SchemaFilterContext context) => Convert.ToInt32(Enum.Parse(context.Type, name));
}
在启动/程序中:
services.AddSwaggerGen(options =>
{
options.SchemaFilter<EnumSchemaFilter>();
});
编辑:对代码进行了一些重构，如果你想看到原始版本，它与其余的答案更相似，请检查编辑的年表。
在。net core 3.1和swagger 5.0.0中:
using System.Linq;
using Microsoft.OpenApi.Any;
using Microsoft.OpenApi.Models;
using Swashbuckle.AspNetCore.SwaggerGen;
namespace WebFramework.Swagger
{
public class EnumSchemaFilter : ISchemaFilter
{
public void Apply(OpenApiSchema schema, SchemaFilterContext context)
{
if (context.Type.IsEnum)
{
var enumValues = schema.Enum.ToArray();
var i = 0;
schema.Enum.Clear();
foreach (var n in Enum.GetNames(context.Type).ToList())
{
schema.Enum.Add(new OpenApiString(n + $" = {((OpenApiPrimitive<int>)enumValues[i]).Value}"));
i++;
}
}
}
}
}
在Startup.cs中:
services.AddSwaggerGen(options =>
{
#region EnumDesc
options.SchemaFilter<EnumSchemaFilter>();
#endregion
});
我想我也有类似的问题。我正在寻找swagger与int ->字符串映射一起生成枚举。API必须接受整型。swagger-ui不那么重要，我真正想要的是代码生成与另一边的“真实”enum(在这种情况下使用改装的android应用程序)。
因此，从我的研究来看，这最终似乎是Swagger使用的OpenAPI规范的一个限制。不能为枚举指定名称和编号。
我发现最好的问题是https://github.com/OAI/OpenAPI-Specification/issues/681，它看起来像“可能很快”，但Swagger将不得不更新，在我的情况下Swashbuckle也是如此。
目前，我的解决方法是实现一个文档过滤器，它查找枚举，并用枚举的内容填充相关的描述。
GlobalConfiguration.Configuration
.EnableSwagger(c =>
{
c.DocumentFilter<SwaggerAddEnumDescriptions>();
//disable this
//c.DescribeAllEnumsAsStrings()
SwaggerAddEnumDescriptions.cs:
using System;
using System.Web.Http.Description;
using Swashbuckle.Swagger;
using System.Collections.Generic;
public class SwaggerAddEnumDescriptions : IDocumentFilter
{
public void Apply(SwaggerDocument swaggerDoc, SchemaRegistry schemaRegistry, IApiExplorer apiExplorer)
{
// add enum descriptions to result models
foreach (KeyValuePair<string, Schema> schemaDictionaryItem in swaggerDoc.definitions)
{
Schema schema = schemaDictionaryItem.Value;
foreach (KeyValuePair<string, Schema> propertyDictionaryItem in schema.properties)
{
Schema property = propertyDictionaryItem.Value;
IList<object> propertyEnums = property.@enum;
if (propertyEnums != null && propertyEnums.Count > 0)
{
property.description += DescribeEnum(propertyEnums);
}
}
}
// add enum descriptions to input parameters
if (swaggerDoc.paths.Count > 0)
{
foreach (PathItem pathItem in swaggerDoc.paths.Values)
{
DescribeEnumParameters(pathItem.parameters);
// head, patch, options, delete left out
List<Operation> possibleParameterisedOperations = new List<Operation> { pathItem.get, pathItem.post, pathItem.put };
possibleParameterisedOperations.FindAll(x => x != null).ForEach(x => DescribeEnumParameters(x.parameters));
}
}
}
private void DescribeEnumParameters(IList<Parameter> parameters)
{
if (parameters != null)
{
foreach (Parameter param in parameters)
{
IList<object> paramEnums = param.@enum;
if (paramEnums != null && paramEnums.Count > 0)
{
param.description += DescribeEnum(paramEnums);
}
}
}
}
private string DescribeEnum(IList<object> enums)
{
List<string> enumDescriptions = new List<string>();
foreach (object enumOption in enums)
{
enumDescriptions.Add(string.Format("{0} = {1}", (int)enumOption, Enum.GetName(enumOption.GetType(), enumOption)));
}
return string.Join(", ", enumDescriptions.ToArray());
}
}
这会在你的swagger-ui上产生如下的结果，这样至少你可以“看到你在做什么”:
