[英]Normal PHP to Codeigniter
我一直在想辦法解決這個問題。 我想將此 HTML PHP 頁面轉換為用 CI 制作的頁面。 這就是我想要得到的(用基本的 php 完成) ,這就是我可以用 Codeigniter 做的事情這是Codeigniter的結果
這是我的意見頁面
<head>
<style>
table td, table th {
padding: 5px;
text-align: center;
}
table th {
background:#3390FF;
}
table {
width: 100%;
border: 2px solid #ccc;
border-collapse: collapse;
}
</style>
</head>
<main role="main" class="col-md-9 ml-sm-auto col-lg-10 px-4">
<div class="d-flex justify-content-between flex-wrap flex-md-nowrap align-items-center pt-3 pb-2 mb-3 border-bottom">
<h1 class="h2">Lista alergeni si alimente</h1>
</div>
<body>
{COUNT}
<h4>Total alimente: {total}</h4>
{/COUNT}
<div style='max-width:500px; margin-right: 1px;'>
<table>
<thead>
<tr>
<th> </th>
{ALERGENI}
<th>{alergen} </th>
{/ALERGENI}
</tr>
<tbody>
{ALIMENTE}
<tr>
<td>{aliment}</td>
<td></td>
</tr>
{/ALIMENTE}
</tbody>
</table>
</div>
</body>
</html>`
這是我的控制器文件
<?php
defined('BASEPATH') OR exit('No direct script access allowed');
class Alergeni_alimente extends CI_Controller {
public function index() {
$lista_alimente = $this->db->query('select alimente.id, alimente.name as alimente_name FROM alimente order by name desc')->result();
$lista_alergeni = $this->db->query('select alergeni.id, alergeni.name FROM alergeni order by name desc')->result();
$content = $this->parser->parse('alergeni_alimente/select', array("ALERGENI" => $lista_alergeni, "ALIMENTE" => $lista_alimente), true);
$TITLE = "Selecteaza alergen pentru aliment ";
$array = array('TITLE' => $TITLE, 'CONTENT' => $content);
$this->parser->parse('TEST', $array, false);
// echo '<pre>';
// print_r($lista_alergeni);
// print_r($lista_alimente);
}
public function selected() {
$data = array(
'alergen_id' => $_POST['alergen_id'],
'aliment_id' => $_POST['aliment_id']
);
$this->db->insert('alimente_alergeni', $data);
redirect('alergeni_alimente');
}
public function list() {
$alergeni_aliment = $this->db->query('select alergen_id, aliment_id, alimente.name as aliment_name, alergeni.name as alergen_name from alimente_alergeni '
. 'inner join alimente on alimente_alergeni.aliment_id = alimente.id '
. 'inner join alergeni on alimente_alergeni.alergen_id = alergeni.id order by alimente.name asc, alergeni.name asc')->result();
$alergeni = $this->db->query('select name as alergen from alergeni')->result();
$alimente = $this->db->query('select name as aliment from alimente')->result();
$totalQuery = $this->db->query("select count(id) as total from alimente")->result();
$content = $this->parser->parse('alergeni_alimente/list', array('ALERGENI_ALIMENT' => $alergeni_aliment, "COUNT" => $totalQuery, 'ALERGENI' => $alergeni, 'ALIMENTE' => $alimente), true);
$TITLE = "Lista alimente si alergenii corespunzatori";
// die();
$array = array('TITLE' => $TITLE, 'CONTENT' => $content);
$this->parser->parse('TEST', $array, false);
}
}
應該怎么做,有人告訴我使用 foreach 語句,但我不確定在哪里。
編輯 1:有人可以幫助我嗎?
我一直在想辦法解決這個問題。 我想將此 HTML PHP 頁面轉換為用 CI 制作的頁面。 這就是我想要得到的(用基本的 php 完成) ,這就是我可以用 Codeigniter 做的事情這是Codeigniter的結果
這是我的意見頁面
<head>
<style>
table td, table th {
padding: 5px;
text-align: center;
}
table th {
background:#3390FF;
}
table {
width: 100%;
border: 2px solid #ccc;
border-collapse: collapse;
}
</style>
</head>
<main role="main" class="col-md-9 ml-sm-auto col-lg-10 px-4">
<div class="d-flex justify-content-between flex-wrap flex-md-nowrap align-items-center pt-3 pb-2 mb-3 border-bottom">
<h1 class="h2">Lista alergeni si alimente</h1>
</div>
<body>
{COUNT}
<h4>Total alimente: {total}</h4>
{/COUNT}
<div style='max-width:500px; margin-right: 1px;'>
<table>
<thead>
<tr>
<th> </th>
{ALERGENI}
<th>{alergen} </th>
{/ALERGENI}
</tr>
<tbody>
{ALIMENTE}
<tr>
<td>{aliment}</td>
<td></td>
</tr>
{/ALIMENTE}
</tbody>
</table>
</div>
</body>
</html>`
這是我的控制器文件
<?php
defined('BASEPATH') OR exit('No direct script access allowed');
class Alergeni_alimente extends CI_Controller {
public function index() {
$lista_alimente = $this->db->query('select alimente.id, alimente.name as alimente_name FROM alimente order by name desc')->result();
$lista_alergeni = $this->db->query('select alergeni.id, alergeni.name FROM alergeni order by name desc')->result();
$content = $this->parser->parse('alergeni_alimente/select', array("ALERGENI" => $lista_alergeni, "ALIMENTE" => $lista_alimente), true);
$TITLE = "Selecteaza alergen pentru aliment ";
$array = array('TITLE' => $TITLE, 'CONTENT' => $content);
$this->parser->parse('TEST', $array, false);
// echo '<pre>';
// print_r($lista_alergeni);
// print_r($lista_alimente);
}
public function selected() {
$data = array(
'alergen_id' => $_POST['alergen_id'],
'aliment_id' => $_POST['aliment_id']
);
$this->db->insert('alimente_alergeni', $data);
redirect('alergeni_alimente');
}
public function list() {
$alergeni_aliment = $this->db->query('select alergen_id, aliment_id, alimente.name as aliment_name, alergeni.name as alergen_name from alimente_alergeni '
. 'inner join alimente on alimente_alergeni.aliment_id = alimente.id '
. 'inner join alergeni on alimente_alergeni.alergen_id = alergeni.id order by alimente.name asc, alergeni.name asc')->result();
$alergeni = $this->db->query('select name as alergen from alergeni')->result();
$alimente = $this->db->query('select name as aliment from alimente')->result();
$totalQuery = $this->db->query("select count(id) as total from alimente")->result();
$content = $this->parser->parse('alergeni_alimente/list', array('ALERGENI_ALIMENT' => $alergeni_aliment, "COUNT" => $totalQuery, 'ALERGENI' => $alergeni, 'ALIMENTE' => $alimente), true);
$TITLE = "Lista alimente si alergenii corespunzatori";
// die();
$array = array('TITLE' => $TITLE, 'CONTENT' => $content);
$this->parser->parse('TEST', $array, false);
}
}
應該怎么做,有人告訴我使用 foreach 語句,但我不確定在哪里。
編輯 1:有人可以幫助我嗎?
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.