php - 从 MySQL 结果和 PHP 为 D3.js 树创建分层 JSON？

Question

我正在尝试使用 PHP 从数据库结果创建以下 JSON(大大简化...):

{
    "name": "Bob",
    "children": [{
            "name": "Ted",
            "children": [{
                "name": "Fred"
            }]
        },
        {
            "name": "Carol",
            "children": [{
                "name": "Harry"
            }]
        },
        {
            "name": "Alice",
            "children": [{
                "name": "Mary"
            }]
        }
    ]
}

数据库表:

Table 'level_1':

level_1_pk| level_1_name
-------------------------
 1 | Bob  


Table 'level_2':

level_2_pk| level_2_name | level_1_fk
-------------------------
 1 | Ted                 | 1
 2 | Carol               | 1
 3 | Alice               | 1


Table 'level_3':

level_3_pk| level_3_name | level_2_fk
-------------------------
 1 | Fred                | 1
 2 | Harry               | 2
 3 | Mary                | 3

代码:

$query = "SELECT * 
FROM level_1
LEFT JOIN level_2
ON level_1.level_1_pk = level_2.level_1_fk";
$result = $connection->query($query);
 while ($row = mysqli_fetch_assoc($result)){
        $data[$row['level_1_name']] [] = array(
            "name" => $row['level_2_name']
            );
    }

echo json_encode($data);

产生:

{"Bob":[{"name":"Ted"},{"name":"Carol"},{"name":"Alice"}]}

问题:

如何获取下一级 level_3，并按照上面定义的 JSON 的要求在 JSON 中包含文本“children”和 level_3 children？

我想如果 JSON 中有更多子项，我将需要 PHP 递归。

SQL

Answer 1

这看起来不像是分层数据的体面设计。考虑另一种方法，例如邻接列表。

解决方案 #1 - MySQL 8 JSON 支持:

对于 MySQL 8，您可以使用 JSON_ARRAYAGG()和 JSON_OBJECT()仅使用 SQL 获取 JSON 结果:

select json_object(
  'name', l1.level_1_name,
  'children', json_arrayagg(json_object('name', l2.level_2_name, 'children', l2.children))
) as json
from level_1 l1
left join (
  select l2.level_2_name
       , l2.level_1_fk
       , json_arrayagg(json_object('name', l3.level_3_name)) as children
  from level_2 l2
  left join level_3 l3 on l3.level_2_fk = l2.level_2_pk
  group by l2.level_2_pk
) l2 on l2.level_1_fk = l1.level_1_pk
group by level_1_pk

结果是:

{"name": "Bob", "children": [{"name": "Ted", "children": [{"name": "Fred"}]}, {"name": "Carol", "children": [{"name": "Harry"}]}, {"name": "Alice", "children": [{"name": "Mary"}]}]}

db-fiddle demo

格式化:

{
  "name": "Bob",
  "children": [
    {
      "name": "Ted",
      "children": [
        {
          "name": "Fred"
        }
      ]
    },
    {
      "name": "Carol",
      "children": [
        {
          "name": "Harry"
        }
      ]
    },
    {
      "name": "Alice",
      "children": [
        {
          "name": "Mary"
        }
      ]
    }
  ]
}

解决方案 #2 - 使用 GROUP_CONCAT() 构造 JSON:

如果名称不包含任何引号字符，您可以使用 GROUP_CONCAT() 在旧版本中手动构建 JSON 字符串:

$query = <<<MySQL
    select concat('{',
      '"name": ', '"', l1.level_1_name, '", ',
      '"children": ', '[', group_concat(
        '{',
        '"name": ', '"', l2.level_2_name, '", ',
        '"children": ', '[', l2.children, ']',
        '}'
      separator ', '), ']'        
    '}') as json
    from level_1 l1
    left join (
      select l2.level_2_name
           , l2.level_1_fk
           , group_concat('{', '"name": ', '"',  l3.level_3_name, '"', '}') as children
      from level_2 l2
      left join level_3 l3 on l3.level_2_fk = l2.level_2_pk
      group by l2.level_2_pk
    ) l2 on l2.level_1_fk = l1.level_1_pk
    group by level_1_pk
MySQL;

结果是一样的(参见 demo )

解决方案 #3 - 使用 PHP 对象构造 nestet 结构:

您还可以编写更简单的 SQL 查询并在 PHP 中构建嵌套结构:

$result = $connection->query("
    select level_1_name as name, null as parent
    from level_1
    union all
    select l2.level_2_name as name, l1.level_1_name as parent
    from level_2 l2
    join level_1 l1 on l1.level_1_pk = l2.level_1_fk
    union all
    select l3.level_3_name as name, l2.level_2_name as parent
    from level_3 l3
    join level_2 l2 on l2.level_2_pk = l3.level_2_fk
");

结果是

name    | parent
----------------
Bob     | null
Ted     | Bob
Carol   | Bob
Alice   | Bob
Fred    | Ted
Harry   | Carol
Mary    | Alice

demo

注意:名称在所有表中应该是唯一的。但我不知道如果可能出现重复，您会期望得到什么结果。

现在将行作为对象保存在由名称索引的数组中:

$data = []
while ($row = $result->fetch_object()) {
    $data[$row->name] = $row;
}

$data 现在将包含

[
    'Bob'   => (object)['name' => 'Bob',   'parent' => NULL],
    'Ted'   => (object)['name' => 'Ted',   'parent' => 'Bob'],
    'Carol' => (object)['name' => 'Carol', 'parent' => 'Bob'],
    'Alice' => (object)['name' => 'Alice', 'parent' => 'Bob'],
    'Fred'  => (object)['name' => 'Fred',  'parent' => 'Ted'],
    'Harry' => (object)['name' => 'Harry', 'parent' => 'Carol'],
    'Mary'  => (object)['name' => 'Mary',  'parent' => 'Alice'],
]

我们现在可以在一个循环中链接节点:

$roots = [];
foreach ($data as $row) {
    if ($row->parent === null) {
        $roots[] = $row;
    } else {
        $data[$row->parent]->children[] = $row;
    }
    unset($row->parent);
}

echo json_encode($roots[0], JSON_PRETTY_PRINT);

结果:

{
    "name": "Bob",
    "children": [
        {
            "name": "Ted",
            "children": [
                {
                    "name": "Fred"
                }
            ]
        },
        {
            "name": "Carol",
            "children": [
                {
                    "name": "Harry"
                }
            ]
        },
        {
            "name": "Alice",
            "children": [
                {
                    "name": "Mary"
                }
            ]
        }
    ]
}

demo

如果可能有多个根节点(level_1_name中的多行)，则使用

json_encode($roots);