如何在同一张表中合并两个SQL查询

Question

使用JSON_CONTAINS方法时，无法使用JOIN获取category_id信息。如何将其转换为单个查询？ 谢谢。

数据库

categories
categories_id   categories_title    categories_parent_id (JSON)     categories_sef_url
1               a                   {"top": "0"}                    a
2               b                   {"top": "0"}                    b
3               c                   {"top": "0"}                    c
4               aa                  {"id": ["1"], "top": "1"}       aa
5               bb                  {"id": ["2"], "top": "2"}       bb
6               cc                  {"id": ["3"], "top": "3"}       cc
7               aaa                 {"id": ["1", "4"], "top": "4"}  aaa
8               bbb                 {"id": ["2", "5"], "top": "5"}  bbb
9               ccc                 {"id": ["3", "6"], "top": "6"}  ccc

SQL

$categories_id = 0;
$sql = "SELECT categories_id from categories where categories_sef_url='aa' ";
$data = $query = $conn->query($sql, PDO::FETCH_ASSOC);
if (!empty($data)) {
    foreach($data as $row) {
        $categories_id = $row->categories_id;
        break;
    }
}   
$sql = "SELECT categories_id, categories_title from categories where **JSON_CONTAINS(categories_parent_id, '{\"id\":\"".$categories_id."\"}')** OR categories_id='".$categories_id."'";
$data = $query = $conn->query($sql, PDO::FETCH_ASSOC);
if (!empty($data)) {
    foreach($data as $row) {
        echo $row->categories_id;
        echo "<br>";
        echo $row->categories_title;
        echo "<hr>";
    }
}

结果

4
aa
7
aaa