我有两个php文件,它们将生成一个jtable插件,当我在不使用代码点火器的情况下运行时,它可以完美地工作。但当我把它放在代码点火器中时,错误就会出现。这是我的文件层次
application/
...
views/
system_admin/
PersonActions.php
view_systemadmin_manageprojects.php
当我运行它时,jtable的基本接口会显示出来,但没有数据和错误。
我在视图文件中有这段代码,它调用了另一个php文件
<script type="text/javascript">
$(document).ready(function () {
//Prepare jTable
$('#PeopleTableContainer').jtable({
title: 'Table of News and Events',
actions: {
listAction: 'PersonActions.php?action=list',
createAction: 'PersonActions.php?action=create',
updateAction: 'PersonActions.php?action=update',
deleteAction: 'PersonActions.php?action=delete'
},
CI框架使用MVC(Model,View,Controller)模式,您必须遵循它,在这里您将直接向PHP PersonActions类发送请求,尝试创建名为personAction的控制器,并将您的操作调用
class PersonActions extends CI_Controller{
function listRecord(){
}
function create(){
}
function update(){
}
function delete(){
}
}
现在是javascript
//Prepare jTable
var base_url = "<?=base_url()?>"; // in you html as global before calling jtable script
$('#PeopleTableContainer').jtable({
title: 'Table of News and Events',
actions: {
listAction: base_url+'PersonActions/list',
createAction: base_url+'PersonActions/create',
updateAction: base_url+'PersonActions/update',
deleteAction: base_url+'PersonActions/delete'
},