Question

我试图从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中拼错了什么吗？

("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);
}

我打印了查询，它仍然返回'21'，我已经排除/或我认为我做了

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 )

Answer 1

有几个问题。看看这里：

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

$notgallery目前是要检查的ID数组。您需要与implode一起重新加入它们，如下所示：

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

此外，您已将gallery_id的NOT IN值包含在引号中。所以事实上你会得到类似的东西：

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

这就像说WHERE gallery_id != '21, 13'。假设您在INT列中使用id，则需要删除$notgallery周围的单引号。如果您使用的是字符串，则可以更改内爆：

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

Answer 2

$ notgallery是一个数组，在你的SQL查询中，你必须有一个用逗号分隔的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());

Answer 3

以上是更好的方式。

$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());

PHP：Mysql（“SELECT WHERE ID NOT”）

3 个答案: