【发布时间】:2014-03-25 02:46:04
【问题描述】:
我最近开始进行单元测试,想知道是否应该编写单元测试以实现 100% 的代码覆盖率?
当我最终编写的单元测试代码多于生产代码时,这似乎是徒劳的。
我正在编写一个 PHP Codeigniter 项目,有时我似乎编写了这么多代码只是为了测试一个小功能。
例如这个单元测试
public function testLogin(){
//setup
$this->CI->load->library("form_validation");
$this->realFormValidation=new $this->CI->form_validation;
$this->CI->form_validation=$this->getMock("CI_Form_validation");
$this->realAuth=new $this->CI->auth;
$this->CI->auth=$this->getMock("Auth",array("logIn"));
$this->CI->auth->expects($this->once())
->method("logIn")
->will($this->returnValue(TRUE));
//test
$this->CI->form_validation->expects($this->once())
->method("run")
->will($this->returnValue(TRUE));
$_POST["login"]=TRUE;
$this->CI->login();
$out = $this->CI->output->get_headers();
//check new header ends with dashboard
$this->assertStringEndsWith("dashboard",$out[0][0]);
//tear down
$this->CI->form_validation=$this->realFormValidation;
$this->CI->auth=$this->realAuth;
}
public function badLoginProvider(){
return array(
array(FALSE,FALSE),
array(TRUE,FALSE)
);
}
/**
* @dataProvider badLoginProvider
*/
public function testBadLogin($formSubmitted,$validationResult){
//setup
$this->CI->load->library("form_validation");
$this->realFormValidation=new $this->CI->form_validation;
$this->CI->form_validation=$this->getMock("CI_Form_validation");
//test
$this->CI->form_validation->expects($this->any())
->method("run")
->will($this->returnValue($validationResult));
$_POST["login"]=$formSubmitted;
$this->CI->login();
//check it went to the login page
$out = output();
$this->assertGreaterThan(0, preg_match('/Login/i', $out));
//tear down
$this->CI->form_validation=$this->realFormValidation;
}
对于这个生产代码
public function login(){
if($this->input->post("login")){
$this->load->library('form_validation');
$username=$this->input->post('username');
$this->form_validation->set_rules('username', 'Username', 'required');
$this->form_validation->set_rules('password', 'Password', "required|callback_userPassCheck[$username]");
if ($this->form_validation->run()===FALSE) {
$this->load->helper("form");
$this->load->view('dashboard/login');
}
else{
$this->load->model('auth');
echo "valid";
$this->auth->logIn($this->input->post('username'),$this->input->post('password'),$this->input->post('remember_me'));
$this->load->helper('url');
redirect('dashboard');
}
}
else{
$this->load->helper("form");
$this->load->view('dashboard/login');
}
}
我哪里错了?
【问题讨论】:
标签: php codeigniter unit-testing