Question

我的网站上有一项功能，让彼此结识的用户可以查看彼此的照片。例如，如果我以Conor登录，并且我想查看Alice's张照片，Conor必须与Alice成为朋友，而Alice必须与Conor成为朋友 - 他们必须是共同的朋友。

我的数据库中有一个名为favourites的表 - 它存储了所有好友请求。假设favourites有两行：

   id: 1
   favourited_who: Alice
   favourited_by: Conor

   id: 2
   favourited_who: Conor
   favourited_by: Alice

他们彼此都有favourited。

请考虑以下代码段：

<?php
$get_favs_q = mysqli_query ($connect, "SELECT * FROM favourites");
        $getting_favs = mysqli_fetch_assoc($get_favs_q);
            $user_favourited = $getting_favs['favourited_who'];
            $user_favourited_by = $getting_favs['favourited_by'];
        /*************************/     
        if ($user == $username  || $user_favourited == $user && $user_favourited_by == $username 
                                || $user_favourited == $username && $user_favourited_by == $user){

            // $user = name in the URL after ?= - As we are on Alice's page .. $user equals Alice
            // $username = session variable for logged in user - $username = Conor

            // If both users have each other favourited, then the code to display images appears here.
        } else {
            echo "  <span style='margin-left: 10px;'> 
                        You and $ufirstname must favourite each other to view each others backstage. 
                    </span>";
        }
?>

我已经涵盖了所有情景，以检查$user是否有利于$username，反之亦然，但是，else语句总是被执行，回应消息而我没有＆＃ 39;不明白为什么。

Answer 1

在支票周围添加括号，如下所示：

if (
    $user == $username  
    || ($user_favourited == $user && $user_favourited_by == $username)
    || ($user_favourited == $username && $user_favourited_by == $user) ){
    // Do your thing
} else {
    // Do the other thing
}

问题在于＆＃34; ands＆＃34;永远都是假的，所以整个陈述总是假的。您可以将您的陈述缩短为以下内容：

A or B and C or D and E

逻辑上与

相同

A or B or D and C and E

从C = !E开始，语句总是错误的。你真正想要的是什么，我上面展示的是：

A or (B and C) or (D and E)

当if期望运行时执行else语句

1 个答案: