相关疑难解决方法(0)

我有以下代码用于注册表单.我想检查电子邮件,看它是否已经登录数据库,如果是,则返回错误.

$name = $_POST['name'];
$email = $_POST['email'];

$result = mysqli_query($connection,"SELECT * FROM users WHERE email='$email'");

if($result=="")
{
echo("email was not found");
}
else
{
    echo("email was found");
}

如果结果变回空洞,我会遇到麻烦.我已经尝试过使用几件事,如果找不到任何东西,就无法弄清楚如何让它正常工作.

我正试图找到最好的方法,看看是否mysqli有结果.我希望代码在有有效查询时执行,如果没有则不执行任何操作.

这是我尝试的代码不起作用:

$result = mysqli_query($con,"SELECT * FROM promoCode where code = '$code'")
    or die("Error in mySql statement!");

if (mysqli_num_rows($result)==1) { echo "yahoo it worked! "; }