php - mysql_fetch_array()期望参数1是资源，给定对象

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Closed 7 years ago.

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PHP: Warning: sort() expects parameter 1 to be array, resource given
我一直有这个奇怪的错误。但当我检查的时候，它看起来很好。

$stuff=mysql_query('SELECT id, name FROM albums');
$albums_list='';
print_r($stuff);
while ($row=mysql_fetch_array($stuff))
{
    print_r($row);
    $albums_list .= "<option value=" . $row[0]['id'] . ">" . $row[0]['name'] . "</option>\n";
}  

print_r($stuff);返回

CI_DB_mysql_result Object ( 
    [conn_id] => Resource id #37 
    [result_id] => Resource id #74 
    [result_array] => Array ( ) 
    [result_object] => Array ( ) 
    [current_row] => 0 
    [num_rows] => 1 
    [row_data] =>
)

Answer 1

这很奇怪，所以mysql_查询返回的是一个对象而不是一个结果。
试试这个：

while ($row=mysql_fetch_array($stuff->result_id))

因为这是对象堆栈中的资源。