Php 检查发送的数据是否与行匹配

我想知道如何检查一行是否与另一行中的内容匹配。例如，如果数据库如下所示：

USERNAME   PASSWORD       EMAIL
Xp10d3     passwordlol    hello@gmail.com
Blurber    password2lol   hello@gmail.com

我想检查用户是否提交了用户名“Xp10d3”，并检查密码是否为passwordlol，提交的电子邮件是否为myemail@gmail.com. 但我不能只检查每个项目，因为可能有两个相同的电子邮件在不同的用户名下（如上所示）。那么，我如何检查发送的数据是否与行匹配？我已经在谷歌上搜索并检查了这个STO： 但这似乎对我不起作用。它说它找不到用户名/密码，即使它已经在数据库中。 我目前的代码是：

<?php
    /* Sends an email to the user and adds the special key to another database */
    $username = $_GET['username']; /* Gets the username that was submitted in the HTML form. */
    $password = $_GET['password']; /* Gets the password that was submitted in the HTML form. */
    $servername = "localhost"; /* MySQL database. Change if needed! Most of the time its not localhost unless you're hosting on your computer. */
    $user = 'usernamelol'; /* MySQL username. Change if needed. */
    $pass = 'passwordlol'; /* MySQL password. Change if needed. */
    $dbname = 'vibemcform'; /* MySQL database name. Change if needed. */

    $bytes = random_bytes(10); /* Randomized code */
    $key = bin2hex($bytes); /* Makes the randomized code */

    $con = new mysqli($servername, $user, $pass, $dbname); /* Connects to the database */
    $query = mysqli_query($con, "SELECT CASE WHEN USERNAME = $username AND PASSWORD = $password THEN 'TRUE ' 
               ELSE 'FALSE' END FROM data");
    /*
    if (mysqli_num_rows($username_query) > 0) {
            echo "Found username!";
    } else {
        echo "Couldn't find that  username!"
        $con -> close();
        exit;
    }
    */
    if ($query === true) {
        echo "Found data in the database!";
    } else {
        echo "Username not found/password incorrect. Please try again!";
    }

    $conn = null;
    echo 'Username submitted: ' . $username . ' Password submitted: ' . $password . ' .'; exit;
?>

---

从（username=“$string”或email=“$string”）和password=“$password”
的表格中选择*。不要忘记散列密码和使用字符串转义/准备语句。谢谢！这个修好了：）谢谢！这很有帮助。发现我的代码中有一些错误。
$query = mysqli_query($con, "SELECT * FROM users WHERE (BINARY username='$string' OR BINARY email='$string') AND BINARY password='$password'");

if(mysqli_num_rows($query) == 1){
    $userCredentialsChecked = TRUE;
}