Question

Google的cAdvisor API提供了如下的JSON输出：

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [],
        'APP_DIRS': True,
        'OPTIONS': {
            'builtins': ['myapp.builtins'],
        },
    },
]

我会将此描述为具有变量/匿名名称的4个相同类型的JSON对象，保存在匿名对象中。

我的第一个想法是做{ /system.slice/docker-13b18253fa70d837e9707a1c28e45a3573e82751f964b66d7c4cbc2256abc266.scope: {}, /system.slice/docker-747f797d19931b4ef33cda0c519f935b592a0b828d16b8cafc350568ab2c1d28.scope: {}, /system.slice/docker-bf947bfabf61cd5168bd599162cf5f5c2ea2350eece1ded018faebf598f7ee5b.scope: {}, /system.slice/docker-e8e02d508400438603151dd462ef036d59fada8239f66be8e64813880b59a77d.scope: { name: "/system.slice/docker-e8e02d508400438603151dd462ef036d59fada8239f66be8e64813880b59a77d.scope", aliases: [...], namespace: "docker", spec: {...}, stats: [...] } }之类的事情，其中：

mapper.readValue(response, Containers.class)

和

public class Containers extends BaseJsonObject {
    @JsonProperty
    public List<Container> containerList;
}

但我能想到的所有变体都会产生相同的结果：public class Container extends BaseJsonObject { @JsonProperty private String name; @JsonProperty public String[] aliases; @JsonProperty private String namespace; @JsonProperty private String spec; @JsonProperty public Stats[] stats; }或com.xyz.Containers@45c7e403[containerList=<null>]的一些排列，com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "/system.slice/docker-13b18253fa70d837e9707a1c28e45a3573e82751f964b66d7c4cbc2256abc266.scope" (class com.xyz.Containers), not marked as ignorable (one known property: "containerList"]) at [Source: java.io.StringReader@3d285d7e; line: 1, column: 97] (through reference chain: com.xyz.Containers["/system.slice/docker-13b18253fa70d837e9707a1c28e45a3573e82751f964b66d7c4cbc2256abc266.scope"])。

我试过了：

ACCEPT_SINGLE_VALUE_AS_ARRAY = false
mapper.readValue(response, Container[].class)
mapper.readValue(response, Containers.class)

以及以下配置：

mapper.readValues(jsonParser, Container.class)
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

如何使用非数组保存的变量/匿名名称解析JSON对象？这叫什么？

Answer 1

您可以按如下方式使用the @JsonAnySetter annotation，并将具有变量名称的对象放入Map<String, Container>的地图中。

以下是一个例子：

public class JacksonVariableNames {

    static final String JSON = "{\n"
            + "  \"a\": {\n"
            + "    \"value\": \"1\"\n"
            + "  },\n"
            + "  \"b\": {\n"
            + "    \"value\": \"2\"\n"
            + "  },\n"
            + "  \"c\": {\n"
            + "    \"value\": \"3\"\n"
            + "  }\n"
            + "}";
    static class Value {
        private final String value;

        @JsonCreator
        Value(@JsonProperty("value") final String value) {this.value = value;}

        @Override
        public String toString() {
            return "Value{" +
                    "value='" + value + '\'' +
                    '}';
        }
    }

    static class Values {
        private final Map<String, Value> values = new HashMap<>();

        @JsonAnySetter
        public void setValue(final String property, final Value value) {
            values.put(property, value);
        }

        @Override
        public String toString() {
            return "Values{" +
                    "values=" + values +
                    '}';
        }
    }
    public static void main(String[] args) throws IOException {
        final ObjectMapper mapper = new ObjectMapper();
        System.out.println(mapper.readValue(JSON, Values.class));

    }
}

输出：

Values{values={a=Value{value='1'}, b=Value{value='2'}, c=Value{value='3'}}}

如何使用Jackson来解析带有可变对象名称的JSON？

1 个答案: