将 属性 类型更改为 Swagger/Swashbuckle 导出的类型
Change property type as exported by Swagger/Swashbuckle
我有一个相当复杂的嵌套对象;请注意,在下面的例子中我大大简化了这个对象。
假设以下示例对象:
public class Result {
public string Name { get; set; }
public IpAddress IpAddress { get; set; }
}
我已经实现了 JsonConverter<IPAddress>
而不是将 Ip 序列化为字符串:
public class IPAddressConverter : JsonConverter<IPAddress>
{
public override IPAddress Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
=> IPAddress.Parse(reader.GetString());
public override void Write(Utf8JsonWriter writer, IPAddress value, JsonSerializerOptions options)
=> writer.WriteStringValue(value.ToString());
}
然后IPAddressConverter
作为AddJsonOptions(...)
方法中的转换器'registered'。这很好 [=70=] 结果为:
{ "Name": "Foo", "IpAddress": "198.51.100.1" }
反之亦然,我的控制器 "understands" IP 地址指定为字符串:
public IEnumerable<Result> FindByIp(IpAddress ip) {
// ...
}
但是,SwashBuckle 将其导出为:
{
"openapi": "3.0.1",
"info": {
"title": "Example",
"version": "v1"
},
"paths": {
"/FindByIp": {
"get": {
"parameters": [
{
"name": "q",
"in": "query",
"schema": {
"type": "array",
"items": {
"type": "string"
}
}
}
],
"responses": {
"200": {
"description": "Success",
"content": {
"application/json": {
"schema": {
"type": "object",
"additionalProperties": {
"$ref": "#/components/schemas/Result"
}
}
}
}
}
}
}
}
},
"components": {
"schemas": {
"AddressFamily": {
"enum": [
0,
1,
2,
3,
4,
5,
6,
6,
7,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19,
21,
22,
23,
24,
25,
26,
28,
29,
65536,
65537,
-1
],
"type": "integer",
"format": "int32"
},
"IPAddress": {
"type": "object",
"properties": {
"addressFamily": {
"$ref": "#/components/schemas/AddressFamily"
},
"scopeId": {
"type": "integer",
"format": "int64"
},
"isIPv6Multicast": {
"type": "boolean",
"readOnly": true
},
"isIPv6LinkLocal": {
"type": "boolean",
"readOnly": true
},
"isIPv6SiteLocal": {
"type": "boolean",
"readOnly": true
},
"isIPv6Teredo": {
"type": "boolean",
"readOnly": true
},
"isIPv4MappedToIPv6": {
"type": "boolean",
"readOnly": true
},
"address": {
"type": "integer",
"format": "int64"
}
},
"additionalProperties": false
},
"Result": {
"type": "object",
"properties": {
"ip": {
"$ref": "#/components/schemas/IPAddress"
},
"name": {
"type": "string",
"nullable": true
}
},
"additionalProperties": false
}
}
}
}
对于更倾向于视觉的人来说,它看起来像:
然而,我想要实现的是:
{
"openapi": "3.0.1",
"info": {
"title": "Example",
"version": "v1"
},
"paths": {
"/FindByIp": {
"get": {
"parameters": [
{
"name": "q",
"in": "query",
"schema": {
"type": "array",
"items": {
"type": "string"
}
}
}
],
"responses": {
"200": {
"description": "Success",
"content": {
"application/json": {
"schema": {
"type": "object",
"additionalProperties": {
"$ref": "#/components/schemas/Result"
}
}
}
}
}
}
}
}
},
"components": {
"schemas": {
"Result": {
"type": "object",
"properties": {
"ip": {
"type": "string",
"nullable": true
},
"name": {
"type": "string",
"nullable": true
}
},
"additionalProperties": false
}
}
}
}
再次可视化:
我希望能够在某些属性上添加注释/属性(所以我查看了 Swashbuckle.AspNetCore.Annotations),但这似乎不可能。
此外,由于该对象相当复杂并且来自第 3 方库,因此我很难在属性上实际添加注释/属性,因为我无法(轻松)更改模型。
我可以求助于 AutoMapper(或类似工具)来创建另一个带有 IP 地址字符串的模型,但这意味着必须对原始模型中的所有对象进行建模。此外,当模型发生变化时,它需要额外的代码和维护。我宁愿以某种方式告诉 Swashbuckle IP 地址(因此,类型 IPAddress
将表示为字符串(传入和传出到我的 API)。我正在寻找选项关于如何在给定限制内以最佳方式完成此操作(最好不要引入要映射到的新模型,最好不要 annotations/attributes 因为我无法轻松访问第 3 方库)。有没有办法注册 "type-converter-something" 让 Swashbuckle 来处理这个?
更新:!
这就是我最终得到的结果:
// This method gets called by the runtime. Use this method to add services to the container.
public void ConfigureServices(IServiceCollection services)
{
services
.AddResponseCompression()
.AddMemoryCache()
.AddControllers()
// etc...
// etc...
// Here's the interesting part:
services.AddSwaggerGen(c =>
{
c.SwaggerDoc("v1", new OpenApiInfo { Title = "Example", Version = "v1" });
c.MapType<IPAddress>(() => new OpenApiSchema { Type = typeof(string).Name });
// ...
});
}
当您转换为非复杂类型时,您应该能够将 MapType
用于此 IP 地址示例:
swagger.MapType<IPAddress>(() => new Schema { Type = "string" });
如果要转换为复杂类型,则需要使用 SchemaFilter。
我有一个相当复杂的嵌套对象;请注意,在下面的例子中我大大简化了这个对象。
假设以下示例对象:
public class Result {
public string Name { get; set; }
public IpAddress IpAddress { get; set; }
}
我已经实现了 JsonConverter<IPAddress>
而不是将 Ip 序列化为字符串:
public class IPAddressConverter : JsonConverter<IPAddress>
{
public override IPAddress Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
=> IPAddress.Parse(reader.GetString());
public override void Write(Utf8JsonWriter writer, IPAddress value, JsonSerializerOptions options)
=> writer.WriteStringValue(value.ToString());
}
然后IPAddressConverter
作为AddJsonOptions(...)
方法中的转换器'registered'。这很好 [=70=] 结果为:
{ "Name": "Foo", "IpAddress": "198.51.100.1" }
反之亦然,我的控制器 "understands" IP 地址指定为字符串:
public IEnumerable<Result> FindByIp(IpAddress ip) {
// ...
}
但是,SwashBuckle 将其导出为:
{
"openapi": "3.0.1",
"info": {
"title": "Example",
"version": "v1"
},
"paths": {
"/FindByIp": {
"get": {
"parameters": [
{
"name": "q",
"in": "query",
"schema": {
"type": "array",
"items": {
"type": "string"
}
}
}
],
"responses": {
"200": {
"description": "Success",
"content": {
"application/json": {
"schema": {
"type": "object",
"additionalProperties": {
"$ref": "#/components/schemas/Result"
}
}
}
}
}
}
}
}
},
"components": {
"schemas": {
"AddressFamily": {
"enum": [
0,
1,
2,
3,
4,
5,
6,
6,
7,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19,
21,
22,
23,
24,
25,
26,
28,
29,
65536,
65537,
-1
],
"type": "integer",
"format": "int32"
},
"IPAddress": {
"type": "object",
"properties": {
"addressFamily": {
"$ref": "#/components/schemas/AddressFamily"
},
"scopeId": {
"type": "integer",
"format": "int64"
},
"isIPv6Multicast": {
"type": "boolean",
"readOnly": true
},
"isIPv6LinkLocal": {
"type": "boolean",
"readOnly": true
},
"isIPv6SiteLocal": {
"type": "boolean",
"readOnly": true
},
"isIPv6Teredo": {
"type": "boolean",
"readOnly": true
},
"isIPv4MappedToIPv6": {
"type": "boolean",
"readOnly": true
},
"address": {
"type": "integer",
"format": "int64"
}
},
"additionalProperties": false
},
"Result": {
"type": "object",
"properties": {
"ip": {
"$ref": "#/components/schemas/IPAddress"
},
"name": {
"type": "string",
"nullable": true
}
},
"additionalProperties": false
}
}
}
}
对于更倾向于视觉的人来说,它看起来像:
然而,我想要实现的是:
{
"openapi": "3.0.1",
"info": {
"title": "Example",
"version": "v1"
},
"paths": {
"/FindByIp": {
"get": {
"parameters": [
{
"name": "q",
"in": "query",
"schema": {
"type": "array",
"items": {
"type": "string"
}
}
}
],
"responses": {
"200": {
"description": "Success",
"content": {
"application/json": {
"schema": {
"type": "object",
"additionalProperties": {
"$ref": "#/components/schemas/Result"
}
}
}
}
}
}
}
}
},
"components": {
"schemas": {
"Result": {
"type": "object",
"properties": {
"ip": {
"type": "string",
"nullable": true
},
"name": {
"type": "string",
"nullable": true
}
},
"additionalProperties": false
}
}
}
}
再次可视化:
我希望能够在某些属性上添加注释/属性(所以我查看了 Swashbuckle.AspNetCore.Annotations),但这似乎不可能。
此外,由于该对象相当复杂并且来自第 3 方库,因此我很难在属性上实际添加注释/属性,因为我无法(轻松)更改模型。
我可以求助于 AutoMapper(或类似工具)来创建另一个带有 IP 地址字符串的模型,但这意味着必须对原始模型中的所有对象进行建模。此外,当模型发生变化时,它需要额外的代码和维护。我宁愿以某种方式告诉 Swashbuckle IP 地址(因此,类型 IPAddress
将表示为字符串(传入和传出到我的 API)。我正在寻找选项关于如何在给定限制内以最佳方式完成此操作(最好不要引入要映射到的新模型,最好不要 annotations/attributes 因为我无法轻松访问第 3 方库)。有没有办法注册 "type-converter-something" 让 Swashbuckle 来处理这个?
更新:!
这就是我最终得到的结果:
// This method gets called by the runtime. Use this method to add services to the container.
public void ConfigureServices(IServiceCollection services)
{
services
.AddResponseCompression()
.AddMemoryCache()
.AddControllers()
// etc...
// etc...
// Here's the interesting part:
services.AddSwaggerGen(c =>
{
c.SwaggerDoc("v1", new OpenApiInfo { Title = "Example", Version = "v1" });
c.MapType<IPAddress>(() => new OpenApiSchema { Type = typeof(string).Name });
// ...
});
}
当您转换为非复杂类型时,您应该能够将 MapType
用于此 IP 地址示例:
swagger.MapType<IPAddress>(() => new Schema { Type = "string" });
如果要转换为复杂类型,则需要使用 SchemaFilter。