我正在使用 codeigniter 2.1.2 做一个 PHP 项目,在 WAMP 上使用 mySql 数据库,我偶然发现了以下问题:
这是我的模型
function loadUserData() {
$sql = "SELECT username FROM user WHERE username = ?";
$query = $this -> db -> query($sql, $this->session->userdata('username'));
if ($query -> num_rows() > 0) {
foreach ($query->result() as $row) {
$data = $row;
}
return $data;
}
}
控制器
function profile_access() {
if ($this -> session -> userdata('is_logged_in')) {
$this->load-> model('profile_model');
$uData['rows']=$this->profile_model->loadUserData();
$this->load->view('profile_view',$uData);
} else {
echo "<script type=\"text/javascript\"> alert(\"You need to provide your credentials first.\");</script>";
redirect('login_controller/index');
}
}
和视图
<?php
foreach($rows as $r){
echo print_r($r);
echo $r[username]->username; //line no. 81 in my code
}
?>
从print_r($r)
视图给我以下错误:
alphaomega1
A PHP Error was encountered
Severity: Notice
Message: Use of undefined constant username - assumed 'username'
Filename: views/profile_view.php
Line Number: 81
A PHP Error was encountered
Severity: Warning
Message: Illegal string offset 'username'
Filename: views/profile_view.php
Line Number: 81
A PHP Error was encountered
Severity: Notice
Message: Trying to get property of non-object
Filename: views/profile_view.php
Line Number: 81
我希望上面的代码只返回数据库中的用户名“alphaomega”。但是会显示它,但我不知道为什么会出现这些错误。将不胜感激一些帮助。谢谢。