php - SQL JOIN 结果不断重复

Question

在我的数据库中，我有 3 个表:user、phone 和 number。他们的结构相应地:

ID | name
---+------
 1 | Joe
 2 | Gin
 3 | Ash


ID |   Brand   | series
---+-----------+--------
 1 |  Samsung  | s7
 2 |  Iphone   | 6s
 3 |  Samsung  | s5


ID |  number   
---+----------
 1 |  77612  
 2 |  34014  
 3 |  98271

我想做的是使用 JOIN 进行选择。这是我的尝试:

$query = "SELECT u.id, p.brand, n.number
          FROM `user` u 
          LEFT OUTER JOIN `phone` p
              ON u.id = p.id
          LEFT OUTER JOIN `number` n 
              ON p.id = n.id
          WHERE u.id = '$selected'";

$sql = mysql_query($query);

if ($sql === FALSE) { 
    die(mysql_error());
}

while ($result = mysql_fetch_row($sql)) {
    $final[] = $result;
    echo '<pre>';
    print_r($final);
    echo '</pre>';
}

其中 $selected 是表单输入的数组数组，允许选择要显示的 ID，例如:

$selected = array(1, 3);

但结果是:

Array (
      [0] => Array (
          [0] => 1
          [1] => Samsung
          [2] => 77612
      )
    )
Array (
      [0] => Array (
          [0] => 1
          [1] => Samsung
          [2] => 77612
      )
      [1] => Array (
          [0] => 3
          [1] => Samsung
          [2] => 98271
      )
)

如果我们设置 $selected = array(1, 2, 3)，输出将与上图相同。我该如何解决这个问题？

Answer 1

$selected =  array(array(1), array(2), array(3));

// make list of id from source array
$selected = call_user_func_array('array_merge', $selected);
$selected = implode(',', $selected); // 1,2,3  

并将 where 子句更改为

WHERE u.id in ($selected)