PHPUnit 测试中包含sesssion怎么处理？

Question

问题描述
使用PHPUnit测试一个方法，该方法中用到了session，然后报错，要先开启session，请问在phpunit中如何开启session

被测方法如下：

    public static function returnJson($arr = [], $type = 0)
    {
        $systemSession = new Zend_Session_Namespace('system');
        $arr['token']  = md5(time());//$systemSession->token;
        header("token: " . $systemSession->token);
        echo json_encode($arr, $type);
        exit;
    }

测试方法如下：

    public function testIsExist()
    {
        $sellerServ = new Application_Service_Api_Seller();
        // 测试无效的参数
        $paramsError = [
            'seller_id'   => 'A2UFATMAM9UT85',
            'customer_id' => 5633
        ];
        $this->assertEquals(true, $sellerServ->isExist($paramsError));
    }

运行phpunit报错信息如下：

Time: 220 ms, Memory: 6.00MB

There was 1 error:

1) ApiSellerTest::testIsExist
Zend_Session_Exception: Session must be started before any output has been sent to the browser; output started in /home/feiffy/demo/KFB/vendor/phpunit/phpunit/src/Util/Printer.php/112

/path/library/Zend/Session.php:451
/path/library/Zend/Session/Namespace.php:143
/path/application/services/Util.php:236
/path/application/services/Util.php:337
/path/application/services/Api/Seller.php:102
/path/tests/apiSellerTest.php:73

ERRORS!

Answer 1

phpunit有一个bootstrap的参数(同样可以写到phpunit.xml配置文件中)，可以指定在测试前预加载一些环境（比如开启session之类的）

不过在cli下开启session没什么用吧。而且你这个类对session有依赖，建议最好用mock的方式来解决