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Php MySQL查询返回非对象?_Php_Mysql - Fatal编程技术网

Php MySQL查询返回非对象?

Php MySQL查询返回非对象?,php,mysql,Php,Mysql,我正在为用户编写代码,以更新用户帐户系统上的密码 这是我所指的代码部分: $checker = mysqli_query($db, "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'"); //echo "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'"; if ($checker == $_SESSION['exp_u

我正在为用户编写代码,以更新用户帐户系统上的密码

这是我所指的代码部分:

$checker = mysqli_query($db, "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'");
//echo "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'";
if ($checker == $_SESSION['exp_user']['userid']) {$check = true;} else {$check = false;}
出于某种原因,
$checker
返回的是对象数组,而不是对象本身

执行
打印($checker)
会产生以下结果:

mysqli_result Object ( [current_field] => 0 [field_count] => 1 [lengths] => [num_rows] => 1 [type] => 0 )
有什么想法吗?

它不返回“对象数组”,只返回一个对象。该对象的类是
mysqli\u result
(如预期),它获得了一些公共字段,如上所示的
num\u rows
。如果要从结果集中读取数据,即使结果集只包含一行和一列,也必须使用类似的获取方法