PHP：Mysql（“SELECT WHERE ID NOT”）

我想从Mysql_query中排除'ID'，但它仍然返回提到的ID。此ID为'21'，但查询返回'21'，这不是我想要的。我在Mysql中拼错了什么？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查询和它仍然返回 '21'，这我已经排除/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)

来源

2013-05-02 Belzelga

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

尝试回显查询（不是结果）。它的内容是什么？ – 2013-05-02 14:52:58

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

有几个问题。看看这里：

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

$notgallery是当前要检查的ID数组。您需要implode加入他们重新走到一起，就像这样：

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

而且，你已经包裹gallery_id的NOT IN引号值。所以实际上你会得到这样的东西：

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

这就像是说WHERE gallery_id != '21, 13'。假定您使用INT s作为id列，则需要删除围绕$notgallery的单引号。如果您使用的是字符串，你可以改变你的破灭：

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

来源

2013-05-02 14:58:36 LeonardChallis

$ notgallery是一个数组，在你的SQL查询中，你必须有一个由逗号分隔的id列表，所以试试：

$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());

来源

2013-05-02 14:55:45 Nagasaki

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

不会'$ notgallery = str_replace（'。'，'，'，$ pic）;'更有意义吗？ – 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());

来源

2013-05-02 16:05:43

PHP：Mysql（“SELECT WHERE ID NOT”）

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