mysql錯誤資源ID＃3？

Question

當我試圖從數據庫中回顯列計數器的值時，我收到資源ID＃3錯誤。我只想得到一個單一的值。任何想法我怎麼能做到這一點？

$Page = $_SERVER['PHP_SELF']; 
$num = preg_replace("/[^0-9]/", '', $Page); 


$query = "SELECT * FROM hitscounter WHERE page='$num';"; 
$res = mysql_query($query); 
if (mysql_num_rows($res) > 0) { 

mysql_query("UPDATE hitscounter SET counter=counter+1 Where page='$num'"); 
$views = mysql_query("SELECT counter FROM hitscounter WHERE page=555"); 
    mysql_fetch_array($views, MYSQL_NUM); 
    mysql_free_result($views); 
    echo $views; 
} 

Answer 1

你應該使用result你從mysql_fetch_array回來，就像這樣：

$result = mysql_fetch_array($views, MYSQL_NUM); 
print_r($result); 

Answer 2

你如果條件應該是這樣的

$result = mysql_fetch_array($res); 
if (count($result) > 0){ 
echo 'do something'; 
}