我有一个非常小的应用程序,它需要一个 id,然后根据该 id 更新一个数据库表。id 是视图的输入。
基本上我的数据库(模型)有getAllProjects()
和approveProject($id)
.
控制器:
<?php if (!defined('BASEPATH')) exit ('No direct script access allowed');
class ApproveProject extends CI_Controller {
public function index () {
$this->loadView();
}
public function getData () {
$this->load->model("db_Projects");
$this->db_Projects->getAllProjects();
}
public function updateDB () {
// how can I get this variable?
$this->db_Projects->approveProject($toApprove);
}
public function loadView() {
$this->load->view("ViewProjectApproval");
}
}
?>
看法:
<html language="en">
<head>
<title>Aprobare Proiect</title>
</head>
<h1> Aprobare Proiect </h1>
<body>
<div id="container">
<?php
if (isset($_POST['projectSubmit']) && ($_POST['projectSubmit'] == "Submit"))
{
$toApprove= $_POST['projectId'];
}
?>
<form action ="updateDB" method="post">
<input type="text" name="projectId">
<input type="submit" name="projectSubmit" value="Submit">
</form>
</div>
</body>
我不知道我的观点是否良好,或者是否应该如此......我怎样才能将 ID 发送到我的控制器?