获取资源ID＃5 printed.dont知道解决方案是什么

Question

$u=$_SESSION['username'];
$p=$_SESSION['password'];
$a = mysql_query("SELECT section FROM users WHERE username ='$u' AND password ='$p'");
echo "$a";

获取资源ID＃5 printed.dont知道问题是什么？

Answer 1

查询返回资源对象。您需要遍历资源以获取每条记录。

取自PHP Docs：

$result = mysql_query($sql);

if (!$result) {
    echo "Could not successfully run query ($sql) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result) == 0) {
    echo "No rows found, nothing to print so am exiting";
    exit;
}

while ($row = mysql_fetch_assoc($result)) {
    echo $row["userid"];
    echo $row["fullname"];
    echo $row["userstatus"];
}

注意： mysql_*函数将为deprecated in PHP 5.5。建议不要编写新代码，因为将来会将其删除。相反，可以是MySQLi或PDO和be a better PHP Developer。