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
}
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。
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”:
如果有人感兴趣，我已经修改了代码的工作
.NET CORE 3和Swagger V5
public class SwaggerAddEnumDescriptions : IDocumentFilter
{
public void Apply(OpenApiDocument swaggerDoc, DocumentFilterContext context)
{
// add enum descriptions to result models
foreach (var property in swaggerDoc.Components.Schemas.Where(x => x.Value?.Enum?.Count > 0))
{
IList<IOpenApiAny> propertyEnums = property.Value.Enum;
if (propertyEnums != null && propertyEnums.Count > 0)
{
property.Value.Description += DescribeEnum(propertyEnums, property.Key);
}
}
// add enum descriptions to input parameters
foreach (var pathItem in swaggerDoc.Paths.Values)
{
DescribeEnumParameters(pathItem.Operations, swaggerDoc);
}
}
private void DescribeEnumParameters(IDictionary<OperationType, OpenApiOperation> operations, OpenApiDocument swaggerDoc)
{
if (operations != null)
{
foreach (var oper in operations)
{
foreach (var param in oper.Value.Parameters)
{
var paramEnum = swaggerDoc.Components.Schemas.FirstOrDefault(x => x.Key == param.Name);
if (paramEnum.Value != null)
{
param.Description += DescribeEnum(paramEnum.Value.Enum, paramEnum.Key);
}
}
}
}
}
private Type GetEnumTypeByName(string enumTypeName)
{
return AppDomain.CurrentDomain
.GetAssemblies()
.SelectMany(x => x.GetTypes())
.FirstOrDefault(x => x.Name == enumTypeName);
}
private string DescribeEnum(IList<IOpenApiAny> enums, string proprtyTypeName)
{
List<string> enumDescriptions = new List<string>();
var enumType = GetEnumTypeByName(proprtyTypeName);
if (enumType == null)
return null;
foreach (IOpenApiAny enumOption in enums)
{
if (enumOption is OpenApiString @string)
{
string enumString = @string.Value;
enumDescriptions.Add(string.Format("{0} = {1}", (int)Enum.Parse(enumType, enumString), enumString));
}
else if (enumOption is OpenApiInteger integer)
{
int enumInt = integer.Value;
enumDescriptions.Add(string.Format("{0} = {1}", enumInt, Enum.GetName(enumType, enumInt)));
}
}
return string.Join(", ", enumDescriptions.ToArray());
}
}
