symfony - 试图从命名空间加载类 "Sm_Image_HandlerController"您是否忘记了另一个命名空间的 "use"语句？

Question

创建类 Tools\MainBundle\Controller\Sm_Image_HandlerController

<?php

namespace Tools\MainBundle\Controller;
class Sm_Image_HandlerController{

    public function test($param) {
        return $param;
    }
}

其他类:

<?php

namespace FoodBundle\Controller;

use FoodBundle\Entity\FoodMaterialMapping;
use Symfony\Component\BrowserKit\Response;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use FoodBundle\Entity\Food;
use Symfony\Component\Validator\Constraints\NotBlank;
use Tools\MainBundle\Controller\Sm_Image_HandlerController;

/**
 * Food controller.
 *
 */
class FoodController extends Controller {
public function indexAction() {
$param="sample";
$sm=new Sm_Image_HandlerController();
$sample=$sm->test($param);
return new \Symfony\Component\HttpFoundation\Response(sample);
}

如果在类 foodController 中运行函数 indexAction返回错误

试图从命名空间“Tools\MainBundle\Controller”加载类“Sm_Image_HandlerController”。您是否忘记了另一个命名空间的“使用”语句？

src/FoodBundle/Controller/FoodController.php 第 23 行

<?php
 public function createAction(Request $request) {
        $param = "sample";
        $sm = new Sm_Image_HandlerController();
        $sample = $sm->test($param);
        return new \Symfony\Component\HttpFoundation\Response(sample);
    }

Answer 1

在 symfony 中你必须使用 CamelCase 命名类和文件名像这样重命名您的 Controller 及其文件

class SmImageHandlerController{

    public function test($param) {
        return $param;
    }
}