如何从DB中创建平面数组的嵌套注释数组

Question

在查询数据库以查找嵌套在闭包表中的注释之后,比如Bill Karwin在这里建议什么是将平面表解析为树的最有效/优雅的方法？,我现在从SQL获得以下数据结构:

"comments": [
            {
                "id": "1",
                "breadcrumbs": "1",
                "body": "Bell pepper melon mung."
            },
            {
                "id": "2",
                "breadcrumbs": "1,2",
                "body": "Pea sprouts green bean."
            },
            {
                "id": "3",
                "breadcrumbs": "1,3",
                "body": "Komatsuna plantain spinach sorrel."
            },
            {
                "id": "4",
                "breadcrumbs": "1,2,4",
                "body": "Rock melon grape parsnip."
            },
            {
                "id": "5",
                "breadcrumbs": "5",
                "body": "Ricebean spring onion grape."
            },
            {
                "id": "6",
                "breadcrumbs": "5,6",
                "body": "Chestnut kohlrabi parsnip daikon."
            }
        ]

使用PHP我想重构这个数据集,所以注释嵌套如下:

"comments": [
            {
                "id": "1",
                "breadcrumbs": "1",
                "body": "Bell pepper melon mung."
                "comments": [
                    {
                        "id": "2",
                        "breadcrumbs": "1,2",
                        "body": "Pea sprouts green bean."
                        "comments": [
                            {
                                "id": "4",
                                "breadcrumbs": "1,2,4",
                                "body": "Rock melon grape parsnip."
                            }
                        ]
                    },
                    {
                        "id": "3",
                        "breadcrumbs": "1,3",
                        "body": "Komatsuna plantain spinach sorrel."
                    }
                ]
            },
            {
                "id": "5",
                "breadcrumbs": "5",
                "body": "Ricebean spring onion grape."
                "comments": [
                    {
                        "id": "6",
                        "breadcrumbs": "5,6",
                        "body": "Chestnut kohlrabi parsnip daikon."
                    }
                ]
            }
        ]

我已经破解了一个解决方案,但它似乎过于复杂,我觉得有一个聪明的解决方案,以优雅和有效的方式做到这一点,但我不知道如何？

Answer 1

假设您将所有数据提取到由“id”索引的数组中：

while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
        $nodes[$row["id"]] = $row;
}

我测试了以下内容，它可以生成您想要的 JSON 输出：

foreach ($nodes as &$node) {
        $parent = array_shift(array_slice(explode(",",$node["breadcrumbs"]), -2, 1));
        if ($parent == $node["id"]) {
                $forest["comments"][] = &$node;
        } else {
                $nodes[$parent]["comments"][] = &$node;
        }
}

print json_encode($forest, JSON_PRETTY_PRINT);