Question

当我运行时，我得到资源#id 3结果。请查看我的代码并帮我解决这个问题......请

<form method="post" action="search.php" name="search_form" onsubmit="return checkCheckBoxes(this);">

         <input type="checkbox" name="search[]" value="qwerty_keypad" id="search1">QWERTY Keypad<br /><br />
         <input type="checkbox" name="search[]" value="touch_screen" id="search2"> Touch Screen<br /><br />
         <input type="checkbox" name="search[]" value="usb" id="search3"> 
         USB Drive<br /><br />
         <input type="checkbox" name="search[]" value="mobile_tracker" id="search4">Mobile Tracker<br /><br />
         <input type="checkbox" name="search[]" value="Backup" id="search5">Phone backup on MMC<br /><br />
         <input type="submit" value="Search" />         </form>

search.php

<?

$host="localhost"; // Host name 
$username="user"; // Mysql username 
$password="password"; // Mysql password 
$db_name="mydb"; // Database name 
$tbl_name="mobile_search"; // Table name

mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");


foreach($_POST['search'] as $search){
    $where[]= "model = '".mysql_real_escape_string($search)."'";
}

$query = "Select model from $tbl_name where ".implode(' OR ',$where);

$result = mysql_query($query);
 echo $result;

?>

Answer 1

这不是mysql结果资源ID的错误。你必须从中获取结果。

while ($row=mysql_fetch_array($result))
{
   print_r($row)// will give complete row
}

Answer 2

阅读本文

mysql_qery（）

函数将仅返回资源ID http://php.net/manual/en/function.mysql-query.php

您需要使用mysql_fetch_array来检索数据

获取资源#id 3错误

2 个答案: