Question

我正在尝试将java接口转换为json架构，但它正在给出NullPointerException

public interface Contributors {

public List<Contributor> contributors();

public interface Contributor {

    public String name();

    public String contributorUrl();

    public List<String> roles();

}

}

编辑2：我得到以下输出：

{"type":"object","$schema":"http://json-schema.org/draft-04/schema#"}

编辑3：

以下是SchemaGeneratorTest的代码

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.github.reinert.jjschema.exception.TypeException;
import com.github.reinert.jjschema.v1.JsonSchemaFactory;
import com.github.reinert.jjschema.v1.JsonSchemaV4Factory;

public class SchemaGeneratorTest {
    private static ObjectMapper mapper = new ObjectMapper();
    public static final String JSON_$SCHEMA_DRAFT4_VALUE = "http://json-schema.org/draft-04/schema#";
    public static final String JSON_$SCHEMA_ELEMENT = "$schema";

    static {
        // required for pretty printing
        mapper.enable(SerializationFeature.INDENT_OUTPUT);
    }

    public static void main(String[] args) throws JsonProcessingException, TypeException {

        JsonSchemaFactory schemaFactory = new JsonSchemaV4Factory();
        schemaFactory.setAutoPutDollarSchema(true);
        JsonNode productSchema = schemaFactory.createSchema(Contributors.class);
        System.out.println(productSchema);
    }
    }

Answer 1

您使用的库仅报告架构中的字段和getter。将您的方法重命名为get：

public interface Contributors {
    public List<Contributor> getContributors();
}

public interface Contributor {
    public String getName();
    public String getContributorUrl();
    public List<String> getRoles();
}

编辑：如果你不能修改接口，你可以使用这个代码来破坏＆＃34; get＆＃34;字符串并让它打印所有方法无论如何。请不要在实际的生产代码中使用它，因为你会给自己带来很多麻烦。

public class Test {

    private static boolean isCorrupted() {
        return "haha".startsWith("get");
    }

    public static void main(String[] args) throws Exception {
        String get = "get";
        Field value = String.class.getDeclaredField("value");
        value.setAccessible(true);
        value.set(get, new char[]{});
        System.out.println(isCorrupted()); // prints true
    }

}

将Java接口转换为Json模式时出错

1 个答案: