Question

我正在使用以下内容来选择一些照片，如何合并'IF EXISTS'以便在没有找到ID的情况下回复一些文字。

$Result = mysql_query("SELECT * FROM photos WHERE UID = '$ID' limit 30")

Answer 1

if (mysql_num_rows($Result)) echo "You have photos!";
else echo "Dude... take some photos!";

Answer 2

如果未找到ID，$ Result将包含0个值。

现在您可以询问，$ Result包含多少值mysql_num_rows()函数，如下所示：

if (mysql_num_rows($Result) == 0) echo "No Photo with $ID found";
else {
     while($photo = mysql_fetch_object($Result)) {
          //Here $photo contains a photo with the UID $ID
     }
}

Answer 3

这就是我最终做的事情。

<?php 
$data = preg_replace ('#[^0-9 ]#i', '', $_POST['data']);
require_once 'db_conx.php';
$Result = mysql_query("SELECT * FROM photos WHERE pid = '$data' limit 30")
 or die (mysql_error());
while($row = mysql_fetch_array($Result)){
$NoPhotos = (empty($row['photo'])) ?  'This Business Profile has no photos, you can send this business a message and ask them to upload some ' : '';
$Photos = (!empty($row['photo'])) ?  '<img width="32%" height="70" src="'.$row['photo'].'" class="ProPhotosID">' : '';
echo '
'.$NoPhotos.''.$Photos.'
';
}
?>

如何检查行中是否存在ID，如果不存在则回显一些文本

3 个答案: