Question

如何使用java对象和Jackson递归创建这个json内容？：

{
    "children": [
        {
            "children": [
                {
                    "name": "cluster",
                    "children": [
                        {
                            "name": "AgglomerativeCluster",
                            "size": 3938
                        },
                        {
                            "name": "TestCLuster",
                            "size": 3938
                        }
                    ]
                }
            ],
            "name": "analytics"
        }
    ],
    "name": "flare"
}

到目前为止，这是我的对象结构，一个名为DendogramVO的父类和一个名为Children的子类，它也有一个子列表：

import java.util.List;

public class DendogramVO {
    private List<Children> children;
    private String name;

    public List<Children> getChildren() {
        return this.children;
    }

    public void setChildren(List<Children> children) {
        this.children = children;
    }

    public String getName() {
        return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

import java.util.List;

public class Children {

    private List<Children> children;
    private String name;

    public List<Children> getChildren() {
        return children;
    }

    public void setChildren(List<Children> children) {
        this.children = children;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

Answer 1

请注意，您的JSON包含未知属性。您需要将ObjectMapper配置为在遇到它们时不要失败（或将它们添加到您的班级）

String s="{ \"children\": [ { \"children\": [ { \"name\": \"cluster\", \"children\": [ { \"name\": \"AgglomerativeCluster\", \"size\": 3938 }, { \"name\": \"TestCLuster\", \"size\": 3938 } ] } ], \"name\": \"analytics\" } ], \"name\": \"flare\"}";
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
Children children = mapper.readValue(s, Children.class);

System.out.println(children.getChildren());

使用简单的Children#toString()打印

[[children = [[children = [[children = null, name = AgglomerativeCluster], [children = null, name = TestCLuster]], name = cluster]], name = analytics]]

如何从json创建java对象

1 个答案: