Question

我正在尝试从数据库中获取数据并检查返回的结果是否为空。这是我的代码。

<?php
require "init1.php";

$jsonObject = $_GET["UserDetails"];
$obj = json_decode($jsonObject);

$email = $obj->Email;
$password = $obj->Password;
$Username = $obj->Username;
$Sex = $obj->Sex;

$timestamp = strtotime($obj->BirthDay);
$BirthDay = date('Y-m-d', $timestamp);

$sql = "select * from Registered_Users where Username='$Username' or Email='$Email'";
$result = mysqli_query($con, $sql);

if (mysqli_num_rows($result) > 0) {

    $myclass = new stdClass();
    $myclass->status = "Not Ok";

    echo json_encode($myclass);

} else {

    try {
        $sql_query = "insert into Registered_Users (Email, Password, BirthDay, Sex, Username) values('$email', '$password', '$BirthDay', '$Sex', '$Username')";
        if (mysqli_query($con, $sql_query)) {
            $obj1 = new stdClass();
            $obj1->status = "Ok";
            echo json_encode($obj1);
        } else {
            $obj1 = new stdClass();
            $obj1->status = "Not Ok";
        }
    } catch (Exception $e) {
        echo $e->getMessage();
    }
}
?>

我能否知道我的上述代码有什么问题，我正在检查数据库中是否存在用户名或电子邮件，如果是，它将打印not ok否则将打印ok。但是当我运行代码时，即使$ result大于它的回显ok，当我第二次运行相同的代码时它会显示NOT OK。有人可以告诉我什么是错的。

谢谢

Answer 1

更改变量名称$ Email并使用$ email

$sql = "select * from Registered_Users where Username='$Username' or Email='$email'";

Php mysqli_query（）错误

1 个答案: