Question

我使用以下代码从带有Code Igniter webapp的MySQL数据库中进行选择：

我的模特

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class m_login extends CI_Model{
function __construct(){
    parent::__construct();
}

public function validate(){
    // grab user input
    $username = $this->security->xss_clean($this->input->post('username'));
    $password = $this->security->xss_clean($this->input->post('password'));

    $password = md5($password);
    // Run the query
    //$array = array('nip' => $this->db->escape_str($username), 'password' => $this->db->escape_str($password));

    //$this->db->select('pegawai.NIP, akun.PASSWORD, akun.ROLE');
    //$this->db->from('pegawai');
    //$this->db->join('akun', 'akun.ID_PEGAWAI = pegawai.ID_PEGAWAI');
    //$this->db->get_where('nip', $username);
    //$this->db->where($array);
    //$this->db->where('PASSWORD', 'd93591bdf7860e1e4ee2fca799911215');
    //$query = $this->db->get();        

    $sql = "SELECT NIP, PASSWORD, ROLE FROM pegawai LEFT JOIN akun ON pegawai.ID_PEGAWAI=akun.ID_PEGAWAI WHERE pegawai.NIP=('".$this->db->escape_str($username)."') AND akun.PASSWORD=('".$this->db->escape_str($password)."')";
    $this->db->query($sql);
    //$query = $this->db->query("SELECT NIP, PASSWORD, ROLE FROM pegawai LEFT JOIN akun ON pegawai.ID_PEGAWAI=akun.ID_PEGAWAI WHERE pegawai.NIP='$username' AND akun.PASSWORD='$password'");
    // Let's check if there are any results
    //var_dump($this->db);
    //var_dump($username);
    if($query->num_rows == 1)
    {
         //If there is a user, then create session data
        $row = $query->row();
        $data = array(
                'role' => $row->role,
                'nip' => $row->nip,
                'validated' => true
                );
        $this->session->set_userdata($data);
        return true;
    }
     //If the previous process did not validate
     //then return false.
    return false;
}
}
?>

我有这样的错误

A PHP Error was encountered 
Severity: Warning
Message: mysql_real_escape_string() expects parameter 2 to be resource, boolean given
Filename: mysql/mysql_driver.php
Line Number: 346
Backtrace:
File: C:\xampp\htdocs\simpeg\application\models\m_login.php
Line: 27
Function: escape_str
File: C:\xampp\htdocs\simpeg\application\controllers\login.php
Line: 26
Function: validate
File: C:\xampp\htdocs\simpeg\index.php
Line: 292
Function: require_once

最有可能的是，这是一个语法错误，但我无法弄清楚我的代码的哪一部分负责。

我还浏览了CodeIgniter用户指南的Queries部分，但在那里没有明确解释。

任何人都可以告诉我我的错误在哪里，以及我正在尝试做什么的正确语法？

消息：mysql_real_escape_string（）期望参数2是资源，给定布尔值文件名：mysql / mysql_driver.php行号：346

0 个答案: