SQL JOIN結果不斷重複

Question

在我的數據庫中，我有3個表格：user，phone和number。他們因此結構：

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 

使用JOIN我想要做的就是選擇。這裏是我的嘗試：

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

哪裏$selected從表單輸入數組的數組，允許選擇什麼ID的顯示，例如：

$selected = array(1, 3); 

但結果是：

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 
    ) 
) 

如果我們設置$selected = array(1, 2, 3)，則輸出將與上面顯示的相同。我怎麼解決這個問題？

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 

和改變where子句

WHERE u.id in ($selected) 

Answer 2

首先，你必須構建你的IDS之間用逗號分隔。爲此我使用的Easiest way to implode() a two-dimensional array?

function implode_r($g, $p) { 
    return is_array($p) ? 
      implode($g, array_map(__FUNCTION__, array_fill(0, count($p), $g), $p)) : 
      $p; 
} 

發現下列函數，那麼你已經使用關鍵字IN改變一點點查詢：

$ids=implode_r(",",$selected); 

$query = "SELECT u.id, p.brand, n.number FROM `user` u 
LEFT OUTER JOIN `phone` p ON u.id = p.id LEFT OUTER JOIN `number` n 
ON p.id = n.id WHERE u.id in ($ids)";