使用 mysqli_num_rows() 时没有任何回显

Question

如果这是一个新手问题，我很抱歉，但我一直在互联网上搜索答案，但找不到任何可以解决我的问题的方法。无论如何，我已经查看了 mysqli_num_rows()（即 https://www.php.net/manual/en/mysqli-result.num-rows.php）上的 php 文档，因为我正在尝试查找列中的行数。我的 table 看起来像这样：

id | follower  |  followee
1    Xp10d3       IiBlurBeriI
2    IiBlurBeriI  Xp10d3

在 id 1 中，table 表明 IiBlurBeriI 有一个 Xp10d3 的追随者，在 id 2 中，table 表明 Xp10d3 有一个 IiBlurBeriI 的追随者。我试图让用户名等于正在查看的个人资料的所有订阅者。无论如何，我使用 mysqli_num_rows() 方法尝试执行此操作，但显然它不起作用。我没有收到任何 MySQL 错误，但是在查看关注者数量时它是空白的，如下所示：

Username
 followers.
 following.

但我希望它看起来像这样：

Username
5 followers.
2 following.

我用来查看个人资料的其余 MySQL 工作正常；我遇到问题的只是 follower/following 系统。我的代码如下：

<?php
    session_start();

    $servername = "localhost"; // Host name 
    $user = "xxxx"; // Mysql username 
    $pass = "xxxx"; // Mysql password 
    $dbname = "xxxx"; // Database name 
    $tbl_name = "forum_question"; // Table name 

    // Connect to server and select databse.
    $conn = new mysqli($servername, $user, $pass, $dbname);
    $userGet = $_GET['username'];
    $userGetSQL = "SELECT USERNAME FROM data WHERE USERNAME='".$userGet."'";
    $result = $conn->query($userGetSQL);
    $userRow = $result->fetch_assoc();

    $pfp = "SELECT PFP FROM data WHERE USERNAME = '".$_GET['username']."'";
    $pfpresult = $conn->query($pfp);
    $pfprow = $pfpresult->fetch_assoc();

    $rank = "SELECT LEVEL FROM data WHERE USERNAME = '".$_GET['username']."'";
    $rresult = $conn->query($rank);

    $followers = "SELECT * FROM subscribers WHERE follower = '".$_GET['username']."'";
    $fresult = $conn->query($followers);
    $fcnt = $fresult->num_rows;

    //echo "FOR TESTING PURPOSES! Followers query: " . $followers . ". Result: " . $fresult . ". num_rows: " . $fcnt . ".";

    $following = "SELECT * FROM subscribers WHERE followee = '".$_GET['username']."'";
    $ffresult = $conn->query($following);
    $ffcnt = $ffresult->num_rows;

    //echo "FOR TESTING PURPOSES! Following query: " . $following . ". Result: " . $ffresult . ". num_rows: " . $ffcnt. ".";

    $desc = "SELECT DESCRIPTON FROM data WHERE USERNAME = '".$_GET['username']."'";
    $descresult = $conn->query($desc);
    $descRow = $descresult->fetch_assoc();

    if (!isset($_SESSION['username']) && empty($_SESSION['username'])) {
        echo 'You are not logged in! Go <a href="main_forum.php">home</a> to login!';
    } else {
?>
<!DOCTYPE HTML>
<html>
<head>
    <link rel="stylesheet" type="text/css" href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css">
    <style>
        body {
            text-align: center;
            font-family: sans-serif;
        }
        .card {
            box-shadow: 0 4px 8px 0 rgba(0, 0, 0, 0.2);
            max-width: 300px;
            margin: auto;
            text-align: center;
        }

        .title {
            color: grey;
            font-size: 18px;
        }

        .msg{
            border: none;
            outline: 0;
            display: inline-block;
            padding: 8px;
            color: white;
            background-color: #000;
            text-align: center;
            cursor: pointer;
            width: 100%;
            font-size: 18px;
        }

        a {
            text-decoration: none;
            font-size: 22px;
            color: black;
        }

        .msg:hover, a:hover {
            opacity: 0.7;
        }
    </style>
</head>
<body>
    <button type="button" style="cursor:pointer">
        <a href="main_forum.php">Home</a>
    </button>
    <br />
    <?php
        $check = mysqli_query($conn, "SELECT * FROM subscribers WHERE follower = '".$_SESSION['username']."' AND followee = '".$_GET['username']."'");
        if (mysqli_num_rows($check) > 0) {
    ?>
            <button type="button" style="cursor:pointer">
                <a href='unfollow.php?username=<?php echo $userGet ?>'>Unfollow</a>
            </button>
    <?php
        } else {
    ?>
            <button type="button" style="cursor:pointer">
                <a href='follow.php?username=<?php echo $userGet ?>'>Follow</a>
            </button>
    <?php
        }
    ?>
    <div class="card">
        <?php
            if ($pfprow['PFP'] == none.png) {
        ?>
            <img src="<?php echo $pfprow['PFP'] ?>" id="pfp" style="width:100%" />
        <?php
        } else {
        ?>
            <img src="pfp/<?php echo $pfprow['PFP'] ?>" id="pfp" style="width:100%" />
        <?php
        }
        ?>
        <h1 id="username"><?php echo $_GET['username'] ?></h2>
        <p id="title"><?php echo $rrow['LEVEL'] ?></p>
        <p><strong><?php echo $fcount ?></strong> followers.</p>
        <p><strong><?php echo $ffcount ?></strong> following.</p>
        <div class="desc">
            <?php
                echo $descRow['DESCRIPTON'];
            ?>
        </div>
        <p><button class="msg"><i class="fa fa-envelope-o" aria-hidden="true"></i> Send Message</button></p>    
    </div>
</body>
</html>
<?php
    }
    exit();
?>

附带说明一下，是的，我知道我的代码容易受到 SQL 注入攻击。我正在尝试学习如何使用准备好的语句（我是 PHP 的新手，但已经使用 HTML+CSS 几年了），以便以后可以更改我的代码.我不相信我有任何 INSERT 语句，所以这应该没问题。

Answer 1

抛开您已经提到的 SQL 注入漏洞，如果您想知道有多少行，例如 follower = 'some_follower'，请执行 COUNT：

SELECT COUNT(1) FROM subscribers WHERE follower = 'some_follower'

因为如果您执行 SELECT * FROM subscribers WHERE follower = 'some_follower'，您要求数据库 return 所有结果，而您只想知道 COUNT。

PHP中的示例：

$result = $conn->query("SELECT COUNT(1) FROM subscribers WHERE follower = 'some_follower'");
$row = $result->fetch_row();

echo '#: ', $row[0];

使用 mysqli_num_rows() 时没有任何回显

Nothing is being echoed when using mysqli_num_rows()

php

mysqli