PHP返回错误消息并返回false

Question

我有一个非常简单的PHP函数来检查登录

function check_login($user, $pass) {
    if(!isset($user) || $user == '') {
        return  'Please enter a valid username';
    }
    else if(!isset($pass) || $pass == '') {
        return  'Please enter a valid password';
    }
    else {
        return 'true';
    }
}

如何在同一时间返回错误消息和false并返回true而不是'true'作为字符串.喜欢,

function check_login($user, $pass) {
    if(!isset($user) || $user == '') {
        return  'Please enter a valid username' //return false;
    }
    else if(!isset($pass) || $pass == '') {
        return  'Please enter a valid password' //return false;
    }
    else {
        return true;
    }
}

所以我可以检查,if(check_login($uname, $pword)){而不是检查像if(check_login($uname, $pword)=='true'){:)

Answer 1

为什么不检查它是否是true,并从那里处理.

$login = check_login('user', 'pass');

if($login === true)
    loginUser(); //login was successful, finalize or whatever
else
    echo $login; //error message

Answer 2

您可以使用的另一种方法是在类中设置错误消息并调用函数以便稍后获取它.

  class exampleClass {
        private $error_message = '';

        function exampleFunction($argument) {

            if($argument == 'lemons') {
                return true;
            } else {
                $this->error_message = 'Argument was not lemons.';
                return false;
            }

        }

        function getErrorMessage() {
            return $this->error_message;
        }

    }

    $example_session = new exampleClass();

    if( ! $example_session->exampleFunction('apples') )
        echo $example_session->getErrorMessage();

Answer 3

我知道这是一篇旧帖子，但是，它得到了很多评论。因此，让我为这里的精彩答案添加另一个答案。

一种解决方案是通过引用传递变量，如 PHP 样式。例子

伪代码：

功能

 //Notice the '&' before the variable name. 
function check_login($user, $pass, &$message) {
    if(!isset($user) || $user == '') {
        $message["status"] = false;  //fail
        $message["message"] ='Please enter a valid username';
    }
    elseif(!isset($pass) || $pass == '') {
           $message["status"] = false;  //fail   
           $message["message"]=  'Please enter a valid password';
    }
    elseif (isset($pass) || $pass != ''){
         $message["status"] = true; //success
         $message["message"]=  'Username and password are valid';
        return true;
    }
       
    return false;//you can return from within the if condition if you wish
}

电话：

/* $message variable will be assigned and populated in the function check_login() */


 
    if(check_login($uname, $pword, $message = []) === true )
    {
       //Do your stuff here
      //@message array is avaialble 
       echo message["message"];
    }
    else
    {
     //means false
     echo message["message"];
    }

Answer 4

为什么不尝试抛出异常

function check_login($user, $pass) {
    if (empty($user)) {
        throw new InvalidArgumentException('Please enter a valid username');
    }
    if (empty($pass)) {
        throw new InvalidArgumentException('Please enter a valid password');
    }
    return true
}

// snip

try {
    check_login($userValue, $passValue);
} catch (Exception $e) {
    // an error occurred
    echo $e->getMessage();
}