PHP將空白信息發送到數據庫

Question

我正在將Unity表單與PHP結合使用，以將數據發送到數據庫，我可以成功地在其中輸入信息並到達數據庫，但是問題是，如果字段為空，它也會發送數據。

我正在嘗試使用isset方法來防止這種情況，但是它似乎沒有用。 有人可以為我再次檢查嗎？

到達數據庫的數據圖像

統一形式的形象

<?php

    if(!isset($_POST['usernamePost'], $_POST['passwordPost'], $_POST['emailPost']))
    {
        echo "This is not working";
    }
    else
    {
    //Enter the data into the database
    $username = filter_var($_POST['usernamePost'], FILTER_SANITIZE_STRING);
    $password = filter_var($_POST['passwordPost'], FILTER_SANITIZE_STRING);
    $email = filter_var($_POST['emailPost'], FILTER_SANITIZE_STRING);

    //variable for connection
    $servername = "localhost";
    $server_username = "root";
    $server_password = "";
    $dbName = "ontrigger_game";

        try
        {
            //connection to the database
            $dbhandle = mysqli_connect($servername, $server_username, $server_password) 
              or die("Unable to connect to MySQL");
            echo "Connected to MySQL<br>";

            //select a database to work with
            $selected = mysqli_select_db($dbhandle, $dbName)
                    or die("Could not select database");

            $sql = "INSERT INTO users (username, password, email)
                VALUES ('".$username."','".$password."','".$email."')";
            if(!mysqli_query($dbhandle, $sql))
            {
              die('Error: ' . mysqli_error());
            }
            echo "1 record added";

            //close the connection
            mysqli_close($dbhandle);

        } 
        catch (Exception $ex) 
        {

        }
    }
?>

Answer 1

您還應該使用empty()方法檢查其是否為空

if (!isset($_POST['usernamePost']) && empty($_POST['usernamePost'] )) {

// error message here

} else {

// What you want to do if not empty and is set.

}

注意：請使用&&運算符，而不要使用逗號