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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pom.xml的配置文件:使用swagger和yaml生成java类
<plugin>
<groupId>io.swagger</groupId>
<artifactId>swagger-codegen-maven-plugin</artifactId>
<version>2.2.3</version>
<executions>
<execution>
<goals>
<goal>generate</goal>
</goals>
<configuration>
<inputSpec>${basedir}/src/main/resources/swagger/project.yaml</inputSpec>
<language>java</language>
<configOptions>
<sourceFolder>src/gen/java/main</sourceFolder>
</configOptions>
</configuration>
</execution>
</executions>
</plugin>
我project.yaml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: com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "attributeA" (class io.swagger.client.model.Parent), not marked as ignorable (2 known properties: "code", "id"])
at [Source: [email protected]; line: 3, column: 17] (through reference chain: io.swagger.client.model.Parent["attributeA"])
我在读几个地方,我需要我的Parent
类的一些注释,如:
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type")
@JsonSubTypes({ @Type(value = ChildA.class, name = "ChildA"),
@Type(value = ChildB.class, 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?
您是否生成这些类用于编写您的服务或这些是为客户端?我有一个需要生成写我的REST服务,请参阅我的问题在这里类:https://stackoverflow.com/q/48799907/1550811 – Learner