php 警告 mysql_fetch_assoc

Question

我正在尝试从 mysql 访问一些信息，但收到警告:mysql_fetch_assoc():提供的参数不是下面第二行代码的有效 MySQL 结果资源，任何帮助将不胜感激。

$musicfiles=getmusicfiles($records['m_id']);
$mus=mysql_fetch_assoc($musicfiles);
for($j=0;$j<2;$j++)
{
 if(file_exists($mus['musicpath']))
 {
  echo '<a href="'.$mus['musicpath'].'">'.$mus['musicname'].'</a>';       
 }
 else
 {
  echo 'Hello world';     
 }
}

function getmusicfiles($m_id)
{
$music="select * from music WHERE itemid=".$s_id;
$result=getQuery($music,$l);
return $result;
}

Answer 1

一般mysql_*函数的用法如下:

$id = 1234;
$query = 'SELECT name, genre FROM sometable WHERE id=' . $id;
// $query is a string with the MySQL query
$resource = mysql_query($query);
// $resource is a *MySQL result resource* - a mere link to the result set
while ($row = mysql_fetch_assoc($resource)) { 
    // $row is an associative array from the result set
    print_r($row);
    // do something with $row
}

如果您向 mysql_fetch_assoc 传递的不是 MySQL 结果资源(无论是字符串、对象还是 bool 值)，该函数将提示它不知道如何处理该参数；这正是您所看到的。

一个常见问题:如果您向 mysql_query 传递某些内容(有效查询字符串除外)，您会收到此警告:

$id = null;
$query = 'SELECT name, genre FROM sometable WHERE id=' . $id;
$res = mysql_query($query); 
// $res === FALSE because the query was invalid
// ( "SELECT name, genre FROM sometable WHERE id=" is not a valid query )
mysql_fetch_assoc($res); 
// Warning: don't know what to do with FALSE, as it's not a MySQL result resource