我是CodeIgniter的新手,当我显示项目列表时遇到问题,我希望每个项目都是一个链接,因此当用户点击该特定项目时,它将显示该项目的详细信息。或者也许就我应该google的内容提出一些建议。我真的不知道应该搜索哪些术语或关键词。
控制器:
<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class Site extends CI_Controller
{
public function index()
{
$this->getListOfAgent();
}
function getListOfAgent()
{
$this->load->model("agentDB_model");
$data['results'] = $this->agentDB_model->getAllAgentInfo();
$this->load->view("viewAgent_view", $data);
}
function userInformation()
{
//how to i get the selected item which the user clicked and show the details accordingly?
}
}
模型
<?php
class agentDB_model extends CI_Model
{
function getAllAgentInfo()
{
$query = $this->db->query("SELECT * FROM user");
return $query->result();
}
function addAgent($newAgent)
{
$this->db->insert("user", $newAgent);
}
}
视图
<!DOCTYPE html>
<html lang="en">
<head>
<title>Welcome</title>
</head>
<body>
<div id="content">
<h1>Home Page</h1>
<p>List Of Agent in database</p>
</div>
<?php
foreach($results as $row)
{
echo "<a href = 'Site/userInformation'>$row->userName</a>";
echo "</br>";
echo "</br>";
}
//$this->load->controller(Site/userInformation);
?>
<div id="footer">
<p>Copyright (c) 2012 basicsite.com</p>
</div>
</body>
</html>
答案 0 :(得分:1)
您需要告诉Site/userInformation
方法您想要哪些记录信息。您可以使用URI segments。
因此,在您的视图中,更改以下行:
echo "<a href = 'Site/userInformation'>$row->userName</a>";
为:
echo "<a href = 'Site/userInformation/$row->userID'>$row->userName</a>";
然后,在您的控制器方法中,将参数添加到方法声明:
function userInformation($userID)
{
// now, you use the model to get the correct record from the db based on
// the $userID and display the information
}