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
}
我想在. net Core应用程序中使用rory_za的答案，但我必须对其进行一些修改才能使其工作。下面是我为。net Core设计的实现。
我还修改了它，这样它就不会假设底层类型是int，并在值之间使用了新的行，以便于阅读。
/// <summary>
/// Add enum value descriptions to Swagger
/// </summary>
public class EnumDocumentFilter : IDocumentFilter {
/// <inheritdoc />
public void Apply(SwaggerDocument swaggerDoc, DocumentFilterContext context) {
// add enum descriptions to result models
foreach (var schemaDictionaryItem in swaggerDoc.Definitions) {
var schema = schemaDictionaryItem.Value;
foreach (var propertyDictionaryItem in schema.Properties) {
var property = propertyDictionaryItem.Value;
var propertyEnums = property.Enum;
if (propertyEnums != null && propertyEnums.Count > 0) {
property.Description += DescribeEnum(propertyEnums);
}
}
}
if (swaggerDoc.Paths.Count <= 0) return;
// add enum descriptions to input parameters
foreach (var pathItem in swaggerDoc.Paths.Values) {
DescribeEnumParameters(pathItem.Parameters);
// head, patch, options, delete left out
var possibleParameterisedOperations = new List<Operation> {pathItem.Get, pathItem.Post, pathItem.Put};
possibleParameterisedOperations.FindAll(x => x != null)
.ForEach(x => DescribeEnumParameters(x.Parameters));
}
}
private static void DescribeEnumParameters(IList<IParameter> parameters) {
if (parameters == null) return;
foreach (var param in parameters) {
if (param is NonBodyParameter nbParam && nbParam.Enum?.Any() == true) {
param.Description += DescribeEnum(nbParam.Enum);
} else if (param.Extensions.ContainsKey("enum") && param.Extensions["enum"] is IList<object> paramEnums &&
paramEnums.Count > 0) {
param.Description += DescribeEnum(paramEnums);
}
}
}
private static string DescribeEnum(IEnumerable<object> enums) {
var enumDescriptions = new List<string>();
Type type = null;
foreach (var enumOption in enums) {
if (type == null) type = enumOption.GetType();
enumDescriptions.Add($"{Convert.ChangeType(enumOption, type.GetEnumUnderlyingType())} = {Enum.GetName(type, enumOption)}");
}
return $"{Environment.NewLine}{string.Join(Environment.NewLine, enumDescriptions)}";
}
}
然后将它添加到Startup.cs中的ConfigureServices方法中:
c.DocumentFilter<EnumDocumentFilter>();
ASP。NET Core 3与Microsoft JSON库(System.Text.Json)
在Startup.cs / ConfigureServices ():
services
.AddControllersWithViews(...) // or AddControllers() in a Web API
.AddJsonOptions(options =>
options.JsonSerializerOptions.Converters.Add(new JsonStringEnumConverter()));
ASP。NET Core 3和Json。NET (Newtonsoft.Json)库
安装Swashbuckle.AspNetCore.Newtonsoft包。
在Startup.cs / ConfigureServices ():
services
.AddControllersWithViews(...)
.AddNewtonsoftJson(options =>
options.SerializerSettings.Converters.Add(new StringEnumConverter()));
// order is vital, this *must* be called *after* AddNewtonsoftJson()
services.AddSwaggerGenNewtonsoftSupport();
ASP。NET Core 2
在Startup.cs / ConfigureServices ():
services
.AddMvc(...)
.AddJsonOptions(options =>
options.SerializerSettings.Converters.Add(new StringEnumConverter()));
Pre-ASP。网络核心
httpConfiguration
.EnableSwagger(c =>
{
c.DescribeAllEnumsAsStrings();
});
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”:
