如何使用java8流在条件比较时递归迭代列表

Question

我有两组json，它们由递归子元素组成。我想要找到该列表中包含"code“的最后一个子代，我想从该json中找到所有"code”值。

Sample1：

{
    "totalSize": 1,
    "data": [
        {
            "level": "sites",
            "children": [
                {
                    "level": "sites",
                    "children": [
                        {
                            "level": "segments",
                            "children": [
                                {
                                    "level": "assets",
                                    "code": "1"
                                },
                                {
                                    "level": "assets",
                                    "code": "2"
                                },
                                {
                                    "level": "assets",
                                    "code": "3"
                                },
                                {
                                    "level": "assets",
                                    "code": "4"
                                },
                                {
                                    "level": "assets",
                                    "code": "5"
                                },
                                {
                                    "level": "assets",
                                    "code": "6"
                                }
                            ]
                        }
                    ]
                }
            ]
        }
    ]
}

Sample2：

{
    "totalSize": 1,
    "data": [
        {
            "level": "sites",
            "children": [
                {
                    "level": "segments",
                    "children": [
                        {
                            "level": "assets",
                            "code": "1"
                        },
                        {
                            "level": "assets",
                            "code": "2"
                        },
                        {
                            "level": "assets",
                            "code": "3"
                        },
                        {
                            "level": "assets",
                            "code": "4"
                        },
                        {
                            "level": "assets",
                            "code": "5"
                        },
                        {
                            "level": "assets",
                            "code": "6"
                        }
                    ]
                }
            ]
        }
    ]
}

实体如下：

public class HierarchyResponse {

    private Integer totalSize;
    private List<Data> data;

}

public class Data {

    private List<Children> children;
    private String level;

}

public class Children {

    private String level;
    private List<Children> children;
}

我试过了，但没有成功：

List<Children> children = response
    .getData()
    .get(0)
    .getChildren()
    .stream().filter(t -> t.getLevel().equalsIgnoreCase("assets"))
    .collect(Collectors.toList());

Answer 1

您可以创建一个递归平铺映射子对象的方法

private static Stream<Children> toStream(Children children) {
    if (children != null) {
        return Stream.concat(Stream.of(children), children.getChildren().stream().flatMap(c -> toStream(c)));
    } else {
        return Stream.empty();
    }
}

并将其用作

List<Children> children = response.getData().stream()
            .flatMap(d -> d.getChildren().stream())
            .flatMap(c -> toStream(c))
            .filter(t -> t.getLevel().equalsIgnoreCase("assets"))
            .collect(Collectors.toList());

例如

Children b1as2 = new Children("assets", "2");
Children b1as1 = new Children("assets", "1");
Children b1seg1 = new Children("segments", "0", List.of(b1as1, b1as2));
Children b1s1 = new Children("sites", "0", List.of(b1seg1));
Children b1 = new Children("sites", "0", List.of(b1s1));

Data data = new Data(List.of(b1));

HierarchyResponse response = new HierarchyResponse(List.of(data));

将打印[Children{level='assets', code='1'}, Children{level='assets', code='2'}]

另一个解决方案未使用流API

private static List<Children> findAssetsChildren(List<Children> children) {
    List<Children> result = new LinkedList<>();
    for(Children ch: children) {
        if(ch.getLevel().equals("assets")) {
            result.add(ch);
        }
        result.addAll(findAssetsChildren(ch.getChildren()));
    }
    return result;
}
//...
List<Children> children = findAssetsChildren(response.getData()
                             .get(0).getChildren());