Php 每次都有错误消息

自从我为我的网站雇佣了一名新的开发人员后，这个错误就开始了。 这完全是错的！当它显然是正确的登录详细信息

你能帮我吗

<?php 
include "global.php";    
?>
<h2>Login</h2>
<?php 
    echo "We currently have <b>" . $usercount . "</b> members, <b>" . $onlinecount . "</b> of which are online. "; 
?>
<br>
<br>
<?php 
    if(isset($_POST["email"])){ 
        $email = $_POST["email"];
        $password = sha1($_POST["password"]);
        $check = mysqli_num_rows($con, "SELECT * FROM Earth WHERE `email`='$email' AND `password`='$password'");
        if($check == 1){
            echo "Logged in!";
        }
        else { 
            echo "WRONG!";
        }
    }
?>
<form action="<?php echo $_SERVER['REQUEST_URI']; ?>" method="post">
    Email <input name="email" placeholder="Email Address" required="" type="text"><br>
    Password <input name="password" placeholder="Password" required="" type="password"><br>
    <input type="reset" value="Start Over">
    <input type="submit" value="Login">
</form>

登录




您根本没有执行查询！
看这里

$check = mysqli_num_rows($con, "SELECT * FROM Earth WHERE `email`='$email' AND `password`='$password'");

您需要使用代码中缺少的mysqli\u query（）执行查询。
为什么要责怪新开发人员？LOL@james你说得对。不要责怪新开发人员，责怪编写了“登录”代码的开发人员。这可能是新开发人员（因为它在过去一直在工作）。那么，mysqli_query（$con，$check）；？不，
$result=mysqli_查询（$con，“您的查询”）$check=mysqli\u num\u行（$result）在这里阅读文档