php-警告：mysql_fetch_object()：提供的参数不是有效的MySQL结果资源

地狱
当我尝试连接并将内容从数据库中拉出时,出现以下错误：

Warning: MysqL_fetch_object(): supplied argument is not a valid MysqL result resource in /home/content/49/5548763/html/matt/download.PHP on line 17

该网站上的其他答案均无效.

这是脚本：

<?PHP

$con = MysqL_connect("XXXX", "name", "password");
if (!$con)
  {
  die('Could not connect: ' . MysqL_error());
  }

$db_selected = MysqL_select_db("nameofdb",$con);

$musictable = "";

$sql = "GET * FROM matt";

$result = MysqL_query($sql,$con);

while($row = MysqL_fetch_object($result)) {

$id = $row->id;

$name = $row->name;

$update = $row->update;

$length = $row->length;

$size = $row->size;

$musictable .= "
  <tr>
    <td width=\"63%\">".$name."</td>
    <td width=\"10%\">".$length." / ".$size."</td>
    <td width=\"10%\"><a href=\"download.PHP?mp3=".$name."\">DOWLOAD</a></td>
    <td width=\"17%\">|||||</td>
  </tr>
  ";
}

?>

解决方法:

那是因为您的查询是错误的.

$sql = "GET * FROM matt";

必须成为

$sql = "SELECT * FROM matt";

被警告的基本措施是

if (!$result)
 die("MysqL error: ". MysqL_error());  

发出查询后.