如何使用PHP和MYSQL从数据库获取5级层次结构用户

Question

我想根据父ID获取子用户ID。我找到了解决办法 如何在php中计算每个级别的15级深度的成员 尝试过的答案 https://stackoverflow.com/a/45535568/23864372 但出现一些错误。

我创建了一个类 -

<?php 
Class Team extends Database { 

 private $dbConnection;

 function __construct($db)
 {
  $this->dbConnection = $db;
 }

 public function getDownline($id, $depth=5) {

        $stack = array($id);
        for($i=1; $i<=$depth; $i++) {

            // create an array of levels, each holding an array of child ids for that level
            $stack[$i] = $this->getChildren($stack[$i-1]);
        }

        return $stack;
    }

    public function countLevel($level) {

         // expects an array of child ids
         settype($level, 'array');

         return sizeof($level);
    }

    private function getChildren($parent_ids = array()) {

        $result = array();
        $placeholders = str_repeat('?,', count($parent_ids) - 1). '?'; 
        $sql="select id from users where pid in ($placeholders)";
        $stmt=$this->dbConnection->prepare($sql);
        $stmt->execute(array($parent_ids));

        while($row=$stmt->fetch()) {

            $results[] = $row->id;
        }

        return $results;
    }
}

我正在使用这样的课程-

$id = 1;
$depth = 2;  // get the counts of his downline, only 2 deep.
$downline_array = $getTeam->getDownline($id, $depth=2);

我收到错误 - 排队

  $placeholders = str_repeat('?,', count($parent_ids) - 1). '?'; 

致命错误：未捕获类型错误：count()：参数 #1 ($value) 必须是 类型为 Countable|array，int 给出于

第二个

警告：PDOStatement::execute()：SQLSTATE[HY093]：无效参数 number：绑定变量的数量与中的标记数量不匹配 线

$sql="select id from users where pid in ($placeholders)";
        $stmt=$this->dbConnection->prepare($sql);

我想获取5个级别的子用户ID。 PHP 版本 8.1

数据库表

superheroes
       -----------
id   parent_id   name
1    0           Steven Rogers
2    1           Bruce Banner
3    1           Wally West
4    2           Peter Parker
5    2           Jay Garrick
6    4           Barry Allen
7    4           Reed Richards

              

Answer 1

有几个小问题：

• $result = array();

  是一个错字，应该是

  $results = array();

• $stack = array($id);

  需要是

  $stack = array([$id]);

  ，因为当您编写

  $stack[$i-1]

  时，它会从该数组中取出一个值，如果没有此更改，则第一次运行代码时，数组中的值只是

  1

  .

• $results[] = $row->id;

  应该是

  $results[] = $row["id"];

  ，因为您是从数据库中以数组而非对象的形式获取行。

这是重写的类：

Class Team extends Database { 

 private $dbConnection;

 function __construct($db)
 {
  $this->dbConnection = $db;
 }

 public function getDownline($id, $depth=5) {

        $stack = array([$id]);
        for($i=1; $i<=$depth; $i++) {

            // create an array of levels, each holding an array of child ids for that level
            $stack[$i] = $this->getChildren($stack[$i-1]);
        }

        return $stack;
    }

    public function countLevel($level) {

         // expects an array of child ids
         settype($level, 'array');

         return sizeof($level);
    }

    private function getChildren($parent_ids = array()) {

        $results = array();
        $placeholders = str_repeat('?,', count($parent_ids) - 1). '?'; 
        $sql="select id from users where pid in ($placeholders)";
        $stmt=$this->dbConnection->prepare($sql);
        $stmt->execute($parent_ids);

        while($row=$stmt->fetch()) {

            $results[] = $row["id"];
        }

        return $results;
    }
}

演示：https://phpize.online/sql/mysql80/431280e176462b7e8001fab0f5b483e6/php/php81/e22e981727ccf23c655b5dbc2160c3d6/