从MySQL结果和PHP for D3.js树创建分层JSON？

Question

我正在尝试使用PHP从数据库结果中创建以下JSON（大大简化了…）：

{
    "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：

如何获得下一个级别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子级？

我想在JSON中有更多子级的情况下，我将需要PHP进行递归。

的SQL

Answer 1

这看起来不像是分层数据的合理设计。考虑另一种方法，例如 邻接表 。

解决方案＃1-MySQL 8 JSON支持：

在MySQL
8中，您可以使用JSON_ARRAYAGG()和JSON_OBJECT()仅通过SQL获得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演示

格式：

{
  "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：

如果名称中不包含引号，则可以使用GROUP_CONCAT()以下版本手动构建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;

结果将是相同的（请参阅demo）

解决方案＃3-使用PHP对象构造嵌套结构：

您还可以编写一个更简单的SQL查询并在PHP中构造嵌套结构：

$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

演示

注意：该名称在所有表中都应该是唯一的。但是，如果有可能重复，我不知道您会期待什么结果。

现在，将行另存为对象，并以名称索引：

$data = []
while ($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'],
]

现在，我们可以在单个循环中链接节点：

$roots = [];
foreach ($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"
                }
            ]
        }
    ]
}

演示

如果可能有多个根节点（中有多行level_1_name），请使用

json_encode($roots);