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我无法理解如何从模型查询中获取数据(单行)到 codeigniter 中的控制器函数中。我搜索了堆栈溢出帖子,但没有运气。我在这样的模型中有一个查询:

function login($email, $sha1Pass) {
    $loginArray = array('email' => $email, 'pw' => $sha1Pass);
    $this->db->select('lid, email, pw');
    $this->db->from('login');
    $this->db->where($loginArray);
    $query = $this->db->get();
    return $query->row();
}

只会返回 1 行[lid, email, pw]

然后我有一个调用这个模型查询的控制器。

$report['row'] = $this->LoginCheck->login($email, $encp); 

我需要的是行中数据库中的 2 个字段,并将它们分配给如下变量:

$lid = $report['lid'];
$email = $report['email'];

lid并且email是数据库中的字段。

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4 回答 4

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您可以像这样访问它们:

$lid = $report['row']->lid;
$email = $report['row']->email;
于 2013-08-15T22:20:58.870 回答
0

尝试这个:

Login_model

public function auth($nm,$pwd)
{
    $this->db->select('uname,password');
    $q1=$this->db->get_where('user_details',array('uname'=>$nm,'password'=>$pwd));
    return $q1->row();
}

//you can call this method from controller like

$data['query']=$this->Log_Model->auth($unm,$pswd);

//And you can create session variable like,

$this->session->set_userdata('uname',$data['query']->uname);
于 2016-04-13T05:18:16.177 回答
0

如果您查看设置此

$lid = $report['lid'];
$email = $report['email'];

然后在模型上添加一些以获取查询

function login($email, $sha1Pass) {
$loginArray = array('email' => $email, 'pw' => $sha1Pass);
$this->db->select('lid, email, pw');
$this->db->from('login');
$this->db->where($loginArray);
$query = $this->db->get();
return $query->row_array();
}

刚刚设置row()row_array()

于 2018-10-03T09:56:36.070 回答
0
  $data = array(
        'username' => $_POST['fullname'],
        'useremail' => $_POST['email'],
        'social_login_id' => $_POST['bq_user_id'],
        'social_login_type' => $loginType,
        'login_status' => 'online',
        'is_active' =>1,
    );
    $this->db->insert($this->TABLE, $data);
于 2018-12-06T06:29:09.220 回答