MYSQL從表1中獲取表2中不存在的記錄

Question

我創建了一個php函數來從sql表中提取記錄subscriptions，我想向mysql_query添加一個條件以忽略表中存在的表subscriptions中的記錄removed_items，這是我的代碼;

function subscriptions_func($user_id, $limit){ 

    $subs = array(); 

    $sub_query = mysql_query(" 
     SELECT `subscriptions`.`fo_id`, `subscriptions`.`for_id`, `picture`.`since`, `picture`.`user_id`, `picture`.`pic_id` 
     FROM `subscriptions` 
     LEFT JOIN `picture` 
     ON `subscriptions`.`fo_id` = `picture`.`user_id` 
     WHERE `subscriptions`.`for_id` = $user_id 
     AND `picture`.`since` > `subscriptions`.`timmp` 
     GROUP BY `subscriptions`.`fo_id` 
     ORDER BY MAX(`picture`.`since_id`) DESC 
     $limit 
     "); 
    while ($sub_row = mysql_fetch_assoc($sub_query)) { 
     $subs [] = array(
      'fo_id'    => $sub_row['fo_id'], 
      'for_id'   => $sub_row['for_id'], 
      'user_id'   => $sub_row['user_id'], 
      'pic_id'   => $sub_row['pic_id'], 
      'since'    => $sub_row['since'] 
     ); 
    } 
    return $subs ; 
} 

我的解決辦法是創建另一個函數從表removed_items獲取記錄，並設置一個PHP條件其中I調用subscriptions_func（）跳過/復位類似於在subscriptions_func（記錄中的記錄），如下面的

$sub = subscriptions_func($user_id); 
foreach($sub as $sub){ 
    $rmv_sub = rmv_items_func($sub[‘pic_id’]); 

    If($rmv_sub[‘pic_id’] != $sub[‘pic_id’]){ 
     echo $sub[‘pic_id’]; 
    } 
} 

將該溶液成功跳過的項目表中的removed_items然而這種解決方案使得存儲在變量$子，這使得在迴盪項木板點陣列中的間隙。

是否有一個條件，我可以添加到函數subscriptions_func（）來削減所有額外的條件和檢查？

Answer 1

假設id是subscriptions和subs_id主鍵是removed_items外鍵，那麼你只需要一個條件添加到WHERE條款。像這樣的東西應該工作：

... 
AND `subscriptions`.id NOT IN (SELECT `removed_items`.subs_id FROM `removed_items`) 
... 

不相關的問題：

• 你的代碼似乎易受SQL injection：使用prepared statement防止這種情況。我們強烈建議您改用Mysqli。