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：
在 MySQL 8 中，您可以使用 JSON_ARRAYAGG() 和 JSON_OBJECT() 仅通过 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，您可以在旧版本中使用 GROUP_CONCAT() 手动构造 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（见demo）
解决方案 #3 - 使用 PHP 对象构建嵌套结构：
你还可以编写一个更简单的 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：名称在所有表中都应该是唯一的。但我不知道你会期待什么结果，如果可以重复的话。
现在将行作为对象保存在按名称索引的数组中：
$data = []
while ($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($roots[0], JSON_PRETTY_PRINT);
结果：
{
"name": "Bob",
"children": [
{
"name": "Ted",
"children": [
{
"name": "Fred"
}
]
},
{
"name": "Carol",
"children": [
{
"name": "Harry"
}
]
},
{
"name": "Alice",
"children": [
{
"name": "Mary"
}
]
}
]
}
demo
如果可能有多个根节点（level_1_name 中有多个行），则使用
json_encode($roots);