Joomla：请求子控制器响应时找不到错误500视图

Question

I'm trying to return the result of a task from a subcontroller with the following url:

index.php?option=com_example&task=subctrl.test&format=json

but I keep getting the 500 View not found Error...

class ExampleControllersSubctrl extends JControllerForm
{
    public function test()
    {
                $result= array("val1","val2");
        echo json_encode($result);
    }
}

I've tried naming the subcontroller file both Subctrl.php & Subctrl.json.php but neither worked. I believe I shouldn't need a view to render the result based on other SO posts I've read but maybe that is incorrect.

This setup will eventually be used to return an Ajax call when I get it working. What am I doing wrong here?

Answer 1

Add an exit statement after the echo statement or Joomla will continue processing the component and will try to call a view. Since no view value was set, no view will be found and the system will redirect to an error page. Full code below:

class ExampleControllerSubctrl extends JControllerForm
{
    public function test()
    {
        $result= array("val1","val2");
        echo json_encode($result);
        exit();
    }
}

Joomla also some other methods that you can use such as call jexit() or JFactory::getApplication()->close() . The general idea is to get the application to stop here. Continuing is a waste.

Also, had to make sure the class name is set right. Middle work should be Controller not Controllers .

Answer 2

The problem is that you're extending JControllerForm which will try and guess the view for your form if one isn't provided.

On Joomla 2.5 you can change JControllerForm to JController and that will resolve the problem.

As you have a JSON controller that Joomla is routing you to via format=json you don't need an exit on your test() method either.