-5
我不知道爲什麼我的表單不起作用,這裏是我的代碼: 因爲無法發佈編輯而添加單詞.............. ................................ PS 所有其他代碼作品完美!表單方法不能正常工作
頁:1:
<button id="Head" onclick="rollHead()">Head</button>
<button id="Tail" onclick="rollTail()">Tail</button>
<script>
function rollTail(){
var die1 = document.getElementById("die1");
var status = document.getElementById("status");
var d1 = Math.floor(Math.random() * 2) + 1;
if(d1 == 1)
{
die1.innerHTML = "You won!";
fliped.innerHTML = "Fliped Tail!";
var ajax = new XMLHttpRequest();
ajax.open('POST','won.php',true);
ajax.send();
}
else if (d1 == 2)
{
die1.innerHTML = "You lose!";
fliped.innerHTML = "Fliped Head!";
var ajax = new XMLHttpRequest();
ajax.open('POST','lose.php',true);
ajax.send();
}
}
function rollHead(){
var die1 = document.getElementById("die1");
var status = document.getElementById("status");
var d1 = Math.floor(Math.random() * 2) + 1;
if(d1 == 1)
{
die1.innerHTML = "You lose!";
fliped.innerHTML = "Fliped Tail!";
var ajax = new XMLHttpRequest();
ajax.open('POST','lose.php',true);
ajax.send();
}
else if (d1 == 2)
{
//alert(document.getElementById('wonorlose').value);
die1.innerHTML = "You won!";
fliped.innerHTML = "Fliped Head!";
var ajax = new XMLHttpRequest();
ajax.open('POST','won.php',true);
ajax.send();
}
}
</script>
<form method="get" action="won.php">
<h2 align="center">
<input type="text" name="wonorlose" value=50>
</h2>
</form>
頁:won.php:
<?php
include_once 'dbconnect.php';
session_start();
if(!isset($_SESSION['user']))
{
header("Location: /manopuslapis/index.php");
}
$res=mysql_query("SELECT * FROM users WHERE user_id=".$_SESSION['user']);
$userRow=mysql_fetch_array($res);
$wonorlose = $_GET["wonorlose"];
echo $wonorlose;
mysql_query("UPDATE `users` SET credits=credits+'$wonorlose' WHERE user_id=".$_SESSION['user']);
?>
您可以更新代碼的工作方式。 – Dave