PHP 仅当两个变量都不为空时，使用逻辑运算符检查空变量才有效

Question

我正在尝试检查我的数据库 table。

它检查提供的用户名是否存在电子邮件地址或密码。如果他们中的任何一个出现在所提供用户名的数据库中，用户将看到一条错误消息并且记录不会被更改。

问题： 但是，出于某种原因，我的检查只有在密码和电子邮件都存在以供记录时才有效（如果电子邮件地址不显示错误消息是空的，但密码不是）。下面是我的代码：

require ('core/dbcon.php');
$user = mysqli_real_escape_string($con, $_GET['user']);
$acc_status = mysqli_real_escape_string($con, $_GET['status']);
if(empty($_POST['form']) === false){
    $usr_email = mysqli_real_escape_string($con, $_POST['email']);
    $pwd = mysqli_real_escape_string($con, $_POST['password']);
    $rpwd = mysqli_real_escape_string($con, $_POST['rpwd']);
    $qry = mysqli_query($con, "SELECT * FROM users WHERE email='$usr_email'") or die(mysqli_error($con));
    $usr_details = mysqli_fetch_assoc($qry);
    if(empty($usr_details['email'])=== true && empty($usr_details['password'])=== true){
        if(mysqli_num_rows($qry) == 0){
            if($pwd && $rpwd && $usr_email){
                if($pwd === $rpwd){
                    $pwd = md5($pwd);

                       $query = mysqli_query($con, "SELECT * FROM users WHERE username='$user' AND email='$usr_email'") or die(mysqli_error($con));
                       if (mysqli_num_rows($query)== 1){
                           //update database table
                       }else{
                            $errors[] = 'Error! Your details were not added to the system due to a technical error. Please contact the Admin.';
                        }           
                    }else{
                        $errors[] = 'Please make sure that the password entered in both fields are the same.';
                    }
                }else{
                    $errors[] = 'Please complete all fields marked with a red asterisk';
                }
            }else{
                $errors[] = 'Error! <b>'.$usr_email.'</b> already existis in our system.';
            }
        }else{
            $errors[] = 'Error! It looks like the username you have entered has been assigned an email address and password already.';
        }
    }

我有一个用户名 'testuser'、电子邮件 'test@email.com' 和密码 'password' 的记录。我的检查工作正常，它显示错误 "Error! It looks like the username you have entered has been assigned an email address and password already." 但是，当我从数据库中删除电子邮件地址时，它以某种方式假定满足以下条件：if(empty($usr_details['email'])=== true && empty($usr_details['password'])=== true){ 当 empty($usr_details['password'])===false。

我尝试将逻辑运算符从 && 更改为 || 但我遇到了同样的问题（因为 OR 应该首先接受一个空变量和一个非空变量）。非常感谢您的帮助。

Answer 1

而不是：

if(empty($usr_details['email'])=== true && empty($usr_details['password'])=== true){

试试这个：

if($usr_details['email'] === NULL || $usr_details['password'] === NULL){

对于空数据库值，NULL 更好。

编辑：

您的查询将 return 为空，因为您只检查 $usr_details['email']，这在数据库中找不到。所以条件会满足。

如果你想正确地得到这个，请尝试使用 username 或 unique id

进行查询

$qry = mysqli_query($con,"select * from user where id=1");

或

$qry = mysqli_query($con,"select * from user where username='username'");

或者您应该在查询中检查 username 和 password：

$qry = mysqli_query($con,"select * from user where email='email' OR password='password'");

Answer 2

我刚刚发现问题出在我的查询上，正如 Lukas Hajdu 所暗示的：

$qry_email = mysqli_query($con, "SELECT email FROM users WHERE email='$usr_email'") or die(mysqli_error($con));

出于某种原因，它仅在数据库中已存在电子邮件地址时才进行 if(empty($usr_details['email'])=== true && empty($usr_details['password'])=== true){ 检查。

为了解决这个问题，我不得不形成另一个查询：

$qry_user = mysqli_query($con, "SELECT * FROM users WHERE username='$user'") or die(mysqli_error($con));
$usr_details = mysqli_fetch_assoc($qry_user);

我不知道为什么会这样，也不知道如何改进我的代码。所以请随时提供建议:)

PHP 仅当两个变量都不为空时，使用逻辑运算符检查空变量才有效

PHP empty variables check with logical operator only works if both variables are not empty

php

if-statement

logical-operators

is-empty