PHP：Mysql（「SELECT WHERE ID NOT」）

我想從Mysql_query中排除'ID'，但它仍然返回提到的ID。此ID爲'21'，但查詢返回'21'，這不是我想要的。我在Mysql中拼錯了什麼？PHP：Mysql（「SELECT WHERE ID NOT」）

("SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ('$notgallery')") or die (mysql_error());

function not_gallery($pic){ 

$pic = $_GET['id']; 
$id = explode(".", $pic); 
$notgallery = $id; 

$notg = mysql_query("SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ('$notgallery')") or die (mysql_error()); 
while($not_row = mysql_fetch_assoc($notg)){ 
    $notgimage[] = array(
     'id' => $not_row['gallery_id'], 
     'user' => $not_row['user_id'], 
     'name' => $not_row['name'], 
     'timestamp' => $not_row['timestamp'], 
     'ext' => $not_row['ext'], 
     'caption' => $not_row['caption'], 

    ); 
} 
print_r($notgimage); 
}

我print_r'ed查詢和它仍然返回 '21'，這我已經排除/或我想我做到

Array ([0] => Array ([id] => 21 [user] => 18 [name] => NotDeojeff [timestamp] => 1367219713 [ext] => jpg [caption] => asd) [1] => Array ([id] => 22 [user] => 18 [name] => NotDeojeff [timestamp] => 1367225648 [ext] => jpg [caption] => Ogre magi)

來源

2013-05-02 Belzelga

你可以print_r'$ _GET ['id']'和'$ notgallery'嗎？ – 2013-05-02 14:52:18

嘗試回顯查詢（不是結果）。它的內容是什麼？ – 2013-05-02 14:52:58

Array（[0] => 21 [1] => jpg）21.jpg – Belzelga 2013-05-02 14:53:35

有幾個問題。看看這裏：

"SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ('$notgallery')"

$notgallery是當前要檢查的ID數組。您需要implode加入他們重新走到一起，就像這樣：

$notgallery = implode(', ', $id);

而且，你已經包裹gallery_id的NOT IN引號值。所以實際上你會得到這樣的東西：

"SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ('21, 13')"

這就像是說WHERE gallery_id != '21, 13'。假定您使用INT s作爲id列，則需要刪除圍繞$notgallery的單引號。如果您使用的是字符串，你可以改變你的破滅：

$notgallery = implode("', '", $id);

來源

2013-05-02 14:58:36 LeonardChallis

$ notgallery是一個數組，在你的SQL查詢中，你必須有一個由逗號分隔的id列表，所以試試：

$pic = $_GET['id']; 
$id = explode(".", $pic); 
$notgallery = $id; 
$notgallery = implode(",", $notgallery); 
$notg = mysql_query("SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ($notgallery)") or die (mysql_error());

來源

2013-05-02 14:55:45 Nagasaki

哇！謝謝。它的工作原理 – Belzelga 2013-05-02 14:57:33

不會'$ notgallery = str_replace（'。'，'，'，$ pic）;'更有意義嗎？ – 2013-05-02 15:03:10

具有上述（POST）更好的方式把它。

$pic = $_GET['id']; 
$id = explode(".", $pic); 
$notgallery = $id; 
$notgallery = implode(",", $notgallery); 
$notg = mysql_query("SELECT * FROM `gallery` WHERE `gallery_id` NOT IN ($notgallery)") or die (mysql_error());

來源

2013-05-02 16:05:43

PHP：Mysql（「SELECT WHERE ID NOT」）

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