PHP 好友请求系统无法正常工作

Question

+----+--------+----------+---------------+---------+
| id | userId | friendId | friendRequest | friends |
+----+--------+----------+---------------+---------+
| 1  | 1      | 3        | true          | false   |
+----+--------+----------+---------------+---------+

这就是我想要的。然而，在页面刷新时，似乎如果您请求任何人作为 friend ，它会以某种方式将每个用户返回为 '$friendRequest' = true 和 '$friends' = false，尽管数据库根本没有列出这些用户......

friend .php

// Connect to USERS table
$userId = $_SESSION['id'];
$con    = mysqli_connect("localhost", "root", "admin123", "master");
$sql    = "SELECT * FROM users ORDER BY id DESC";
$result = mysqli_query($con, $sql);

// Connect to FRIENDS table
$sqli          = $con->query("SELECT * FROM friends");
$data          = $sqli->fetch_array();
$friendRequest = $data['friendRequest'];
$friends       = $data['friends'];

// Build USERS table
while ($row = mysqli_fetch_array($result)) {
    $profile   = $row['profile'];
    $id        = $row['id'];
    $firstName = $row['firstName'];
    $lastName  = $row['lastName'];
    $city      = $row['city'];
    $state     = $row['state'];
    $bio       = $row['bio'];

    // Exclude user profile, current friends, and active friend requests

    if ($id !== $userId && $friends !== "true" && $friendRequest !== "true") {

        echo "
   <div class='card m-0'>
      <div class='text-center align-middle bg-white pt-3 pb-3'>
         <img class='circle border mx-auto' style='width: 100px; height: 100px; object-fit: cover;' src='profile_pics/" . $profile . "'>
         <h4 class='text-black font-weight-bolder'>" . $firstName . " " . $lastName . "</h4>
      <div class='h7'><i class='fas fa-map-marker-alt text-danger'></i> " . $city . ", " . $state . "</div>
         <form action='friends.php' method='post'>
            <a href='request.php?id=" . $id . "' class='btn btn-primary mx-auto mt-3'>Send Friend Request</a>
         </form>
      </div>
   </div>";

    } else {
        echo "Friend request: " . $friendRequest . "<br>Friends? " . $friends . "<br>";
    }
}

请求.php

<?php
     require "functions.php";
     require "logincheck.php";

  $userId = $_SESSION['id'];
  $friendId = $_GET['id'];
  $con = mysqli_connect("localhost", "root", "", "master");
  $sqli = ("INSERT INTO friends (userId, friendId, friendRequest, friends) VALUES ('$userId', '$friendId', 'true', 'false')");
  mysqli_query($con, $sqli);

  header("Location: friends.php");

?>

Answer 1

在获取 friends 表的第一行并设置变量 $friendRequest 和 $friends 后，它们表示用户 1 之间的关系和用户 3。当您开始循环所有用户时，您永远不会更改这些值，因此即使您检查例如，它们仍然代表这些用户之间的关系。用户2。

我想您想要的是选择特定用户的所有(待处理)好友请求(SELECT * FROM Friends WHERE userId = ? - 确保使用准备好的语句!)，然后在循环中检查当前循环的用户是否出现在 friends 查询的结果中。如果是，您已经向他发送了好友请求。