php - MYSQL 查询返回“资源 id#12”而不是它应返回的数值

Question

不知道为什么，但这返回了错误的值。我正在取回此资源 ID #12，而不是我正在寻找的数值“1”..

执行此操作的代码是:

$type = "SELECT account_type from user_attribs WHERE username = '$username'";
$usertype= mysql_query($type);

所以我将其更改为:

$type = "SELECT account_type from user_attribs WHERE username = '$username'";
$type_again = mysql_query($type);
$usertype = mysql_fetch_row($type_again);

现在它只给我单词数组。我对此完全迷失了。帮忙？!

编辑::

当前使用的代码是:

  $username= 'lmfsthefounder';
  $type = "SELECT account_type from user_attribs WHERE username='lmfsthefounder';";
  $type_again = mysql_query($type);
  $row = mysql_fetch_row($type_again);
  $usertype = $row['account_type'];
  echo $usertype;

返回如下:主页登录注册用户类型是

(这应该在我的导航栏中显示“Usertype is 1”)

Answer 1

你就快到了。您有包含 MySQL 结果的行，这就是 mysql_fetch_row() 返回的内容。将其更改为 mysql_fetch_assoc()这使您的代码更具可读性。然后，您只需要您要查找的特定列值，您可以通过使用其名称作为数组键来访问该列值:

$type = "SELECT account_type from user_attribs WHERE username = '$username'";
$type_again = mysql_query($type);
$row = mysql_fetch_assoc($type_again);
echo $row['account_type'];

Please, don't use mysql_* functions in new code 。它们不再维护and are officially deprecated 。请参阅red box ？了解 prepared statements相反，并使用 PDO或MySQLi -this article将帮助您决定哪个。如果您选择 PDO，here is a good tutorial .