使用数据作为树视图过滤 ArrayList

Question

我有一个类 Detail.java 带有一些属性 id、security、description、items。 Items 是一个 ArrayList 的详细信息

public class Detail {

    private int id;
    private boolean security;
    private String description;
    private List<Detail> items;

    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public boolean isSecurity() {
        return security;
    }
    public void setSecurity(boolean security) {
        this.security = security;
    }
    public String getDescription() {
        return description;
    }
    public void setDescription(String description) {
        this.description = description;
    }
    public List<Detail> getItems() {
        return items;
    }
    public void setItems(List<Detail> items) {
        this.items = items;
    }
}

这个 ArrayList (treeview) 和我的数据

[
    {
        "id": 1,
        "security": true,
        "description": "description 1",
        "items": [
            {
                "id": 11,
                "security": true,
                "description": "description 11"
            },
            {
                "id": 12,
                "security": false,
                "description": "description 12",
                "items": [
                    {
                        "id": 121,
                        "security": true,
                        "description": "description 121"
                    },
                    {
                        "id": 122,
                        "security": false,
                        "description": "description 122"
                    }
                ]
            },
            {
                "id": 13,
                "security": true,
                "description": "description 13"
            },
            {
                "id": 14,
                "security": false,
                "description": "description 14"
            }
        ]
    },
    {
        "id": 2,
        "security": true,
        "description": "description 2",
        "items": [
            {
                "id": 21,
                "security": true,
                "description": "description 21"
            },
            {
                "id": 22,
                "security": false,
                "description": "description 22"
            },
            {
                "id": 23,
                "security": true,
                "description": "description 23"
            },
            {
                "id": 24,
                "security": false,
                "description": "description 24"
            }
        ]
    }

]

我想通过过滤所有安全性为真的节点来过滤或创建这个ArrayList

最好的方法是什么？使用迭代器？用第一个创建一个新的arrayList 和克隆对象？ 如何管理物品？

感谢您的帮助

Answer 1

最近看到这个问题，试了一下。这个递归解决方案似乎对我有用：

public class Main {

// ArrayList to popluate with elements
private static List<Detail> myDetails2 = new ArrayList<>();

public static void main(String[] args) {

    List<Detail> myDetails = new ArrayList<>();

    // Building up some sample data
    Detail detail = new Detail();
    detail.setId(1);
    detail.setSecurity(true);
    detail.setItems(Collections.emptyList());

    Detail detail5 = new Detail();
    detail5.setId(6);
    detail5.setSecurity(true);
    detail5.setItems(Collections.emptyList());

    Detail detail4 = new Detail();
    detail4.setId(5);
    detail4.setSecurity(true);
    detail4.setItems(Arrays.asList(detail5));

    Detail detail3 = new Detail();
    detail3.setId(4);
    detail3.setSecurity(true);
    detail3.setItems(Arrays.asList(detail4));

    Detail detail2 = new Detail();
    detail2.setId(3);
    detail2.setSecurity(true);
    detail2.setItems(Arrays.asList(detail3));

    Detail detail1 = new Detail();
    detail1.setId(2);
    detail1.setSecurity(false);
    detail1.setItems(Arrays.asList(detail2));

    myDetails.add(detail);
    myDetails.add(detail1);

    constructDetailsListWithSecurity(myDetails);
    // Printing out the id of the elements that were added to the filtered ArrayList
    myDetails2.forEach(d -> System.out.println("id: " + d.getId()));
}

public static void constructDetailsListWithSecurity(List<Detail> myDetails) {

    if (myDetails.isEmpty()) {
        return;
    }

    for (Detail detail : myDetails) {
        if (detail.isSecurity()) {
            myDetails2.add(detail);
        }
        if (detail.getItems() != null && !detail.getItems().isEmpty()) {
            List<Detail> items = detail.getItems();
            constructDetailsListWithSecurity(items);
        }
    }   
}

}

我基本上只使用安全设置为 true 的元素填充 ArrayList（我希望我能理解这一点，否则我们需要稍微更改逻辑）。对于此解决方案，深度无关紧要。这意味着它仍然可以处理 Detail 对象具有 Details 列表的情况，该列表又具有 Details 列表等等。