[英]Class 'JFactory' not found
I've been creating some modules in Joomla using Jumi. 我一直在使用Jumi在Joomla中创建一些模块。 So I can write any php/javascript code and create a Jumi module that I can display where I want.
因此,我可以编写任何php / javascript代码并创建一个Jumi模块,该模块可以显示在所需的位置。
I've been doing this for a while without problems but now that I'm trying some AJAX development with Jquery I'm getting this error: 我这样做已经有一段时间了,但是现在我正在尝试使用Jquery进行一些AJAX开发,但是出现了这个错误:
Class 'JFactory' not found in api.php
So I have a PHP file with the jQuery code: 所以我有一个带有jQuery代码的PHP文件:
$(function() {
$.ajax({
url: 'ajax_dashboard/api.php', //the script to call to get data
data: "",
dataType: 'json', //data format
success: function(data) //on recieve of reply
{
var id = data[0]; //get id
var vname = data[1]; //get name
$('#output').append("<b>id: </b>"+id+"<b> name: </b>"+vname)
.append("<hr />"); //Set output element html
}
});
});
As you can see it calls the api.php script to do some server processing. 如您所见,它调用api.php脚本进行一些服务器处理。 This file has a number of joomla calls like:
该文件有许多joomla调用,例如:
$user = &JFactory::getUser();
So why in this case do I not have the Joomla framework available? 那么为什么在这种情况下我没有Joomla框架呢?
The problem is that your Ajax call ends up in a file out of the Joomla "platform". 问题是您的Ajax调用最终会出现在Joomla“平台”之外的文件中。 The proper way to do that, if possible, is to make the ajax call something like:
如果可能的话,正确的方法是使ajax调用类似以下内容:
index.php?option=yourcomponent&controller=xxx&task=yyy 的index.php?选择= yourcomponent&控制器= XXX&任务= YYY
(it means you should have a component "mycomponent" and a controller "xxx" inside that component ) Then the controller should be responsible to handle the ajax call and send a response. (这意味着您应该在该组件内部有一个组件“ mycomponent”和一个控制器“ xxx”),然后该控制器应负责处理ajax调用并发送响应。 You can return json-encoded response for example or anything you need.
您可以返回json编码的响应,例如您需要的任何内容。
I hope it helped 希望对您有所帮助
I use this to solve the problem. 我用它来解决问题。 I get the variable while I'm in the joomla framework.
我在joomla框架中时得到了变量。 Then I pass the User_Name variable in my ajax call...
然后我在ajax调用中传递User_Name变量...
Hope this helps 希望这可以帮助
<script type="text/javascript">
var User_Name = '<?php $user =& JFactory::getUser(); $User_Name = $user->username; echo $User_Name; ?>';
</script>
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