我想用maven插件swagger-codegen-maven-plugin版本2.2.3生成我的Java类。在这里我的的配置文件：使用swagger和yaml生成java类
swagger-codegen-maven-plugin
2.2.3
generate
${basedir}/src/main/resources/swagger/
java
src/gen/java/main
我文件包含此：产生
definitions:
Parent:
type: "object"
discriminator: "type"
required:
- type
properties:
id:
type: "integer"
format: "int64"
code:
type: "string"
ChildA:
allOf:
- $ref: "#/definitions/Parent"
- properties:
attributeA:
type: "string"
ChildB:
allOf:
- $ref: "#/definitions/Parent"
- properties:
attributeB:
type: "string"
所有3类，然后我想用网络来创建ChildA或ChildB服务。所以，我的方法是：
@POST
public Response createChild(@WebParam Parent parent) {
...
}
使用邮差，我把下面的JSON，以便创造一个ChildA实例：
{
"code": "child-a",
"attributeA": "value"
}
以下例外情况：
Caused by: : Unrecognized field "attributeA" (class ), not marked as ignorable (2 known properties: "code", "id"])
at [Source: [email protected]; line: 3, column: 17] (through reference chain: ["attributeA"])
我在读几个地方，我需要我的Parent类的一些注释，如：
@JsonTypeInfo(use = , include = , property = "type")
@JsonSubTypes({ @Type(value = , name = "ChildA"),
@Type(value = , name = "ChildB") })
但我不知道如何修改我的YAML文件中添加的这些注释。有人可以帮助我吗？
2017-10-06
cheb1k4