如何检查行中是否存在ID，如果不存在则回显一些文本[关闭]

问题描述：

I'm using the following to select some photo, how do I incorporate 'IF EXISTS' so i can echo some text if no ID is found.

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

我正在使用以下内容选择一些照片，如何合并“IF EXISTS”以便我能回应一些如果没有找到ID，则为文本。 p>

  $ Result = mysql_query（“SELECT * FROM photos WHERE UID ='$ ID'limit 30”）
  code>   pre> 
  div>

答

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

答

When no ID is found, the $Result will contain 0 values.

Now you can ask, how many values $Result contains with mysql_num_rows() function like this:

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
     }
}

答

Here's what i ended up doing.

<?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[关闭]

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