Question

我正在尝试创建一个系统，允许我为自己的游戏创建一种日常使用礼物。如果玩家几乎每天都登录游戏（很显然是在当天的重置时间之后），那么他每天的首次访问都会获得奖金。尽管是PHP和MySql的新手，但我还是尝试编写一些代码。第一部分似乎正常工作，但是编写第二部分代码时遇到问题。

<?php
// dati accesso database
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dailybonus";

// Create connection
$mysqli = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($mysqli->connect_error) {
    die("Connessione fallita");
}

// Check if the player is already present in the database, or if it is his first ever login
$check = $mysqli->query('SELECT * FROM bonus WHERE idgiocatore = "'.$_GET["idgiocatore"].'"');
// If it's your first login
if($check->num_rows == 0) {
    echo "Primo login del giocatore!<br>";
    // I insert his id and the current date / time in the database
    $now = new DateTime();
    $update = $mysqli->query('INSERT INTO bonus (idgiocatore,ultimoaccesso) VALUES ("'.$_GET["idgiocatore"].'","'.date_format($now, 'd-m-Y H:i:s').'")'); 
    // If there are no errors I return "true" for the bonus assignment
    if ($update) {
        echo "true";
    // Otherwise I return the error
    } else { die("ERRORE");}
} else {
    // Otherwise if the player is already present in the database

    $now = new DateTime();
    $resetTime = new DateTime();
    date_time_set($resetTime, 12, 00, 00);
    $a = $mysqli->query('SELECT ultimoaccesso FROM bonus WHERE idgiocatore = "'.$_GET["idgiocatore"].'"');
    $lastLogin = new DateTime((string)$a);
    echo "LastLogin: " . date_format($lastLogin, 'd-m-Y H:i:s') . "<br>";
    echo "Reset time: " . date_format($resetTime, 'd-m-Y H:i:s') . "<br>";
    echo "time: " . date_format($now, 'H') . "<br>";

    if ($resetTime < $now) {
        echo "ok1<br>";
        if ($resetTime < $lastLogin) {
            echo "ok2";
        }
    }



    $update = $mysqli->query('UPDATE bonus SET ultimoaccesso=NOW() WHERE idgiocatore="'.$_GET["idgiocatore"].'"');
    if ($update) {
         echo "Aggiornato";
     } else { die("ERRORE");}
}

$mysqli->close(); 
?>

当前我无法比较日期，因为两个日期之一不是字符串，而且我不知道如何转换。显然，我不确定我使用的是获得所需结果的最简单/最快的方法。

你能帮我一下吗？

比较php和mysqli日期

0 个答案: