Ich möchte ein Karten-Matching-Spiel erstellen, aber ich habe ein Problem mit den Bildern, die versteckt werden sollen. Wenn ich auf eine Karte klicke, wird der Pfad des Bildes anstelle des tatsächlichen Bildes angezeigt.Erstellen eines Karten-Matching-Spiels
Hier sind alle Codes, die ich geschrieben habe:
<!DOCTYPE html>
<html>
<head>
<title></title>
<meta charset="utf-8" />
<style type="text/css">
div#card_board
{
background: #ccc;
border: #999 1px solid;
width: 710px;
height: 600px;
padding: 24px;
margin: 0px auto;
}
div#card_board > div
{
background: url(cardQtion.jpg) no-repeat;
background-size: 100%;
border: #000 1px solid;
width: 114px;
height: 132px;
float: left;
margin: 10px;
padding: 20px;
font-size: 64px;
cursor: pointer;
text-align: center;
}
</style>
<script>
var cardArray = new Array();
cardArray[0] = new Image();
cardArray[0].src = 'cardA.jpg';
cardArray[1] = new Image();
cardArray[1].src = 'cardA.jpg';
cardArray[2] = new Image();
cardArray[2].src = 'cardB.jpg';
cardArray[3] = new Image();
cardArray[3].src = 'cardB.jpg';
cardArray[4] = new Image();
cardArray[4].src = 'cardC.jpg';
cardArray[5] = new Image();
cardArray[5].src = 'cardC.jpg';
cardArray[6] = new Image();
cardArray[6].src = 'cardD.jpg';
cardArray[7] = new Image();
cardArray[7].src = 'cardD.jpg';
cardArray[8] = new Image();
cardArray[8].src = 'cardE.jpg';
cardArray[9] = new Image();
cardArray[9].src = 'cardE.jpg';
cardArray[10] = new Image();
cardArray[10].src = 'cardF.jpg';
cardArray[11] = new Image();
cardArray[11].src = 'cardF.jpg';
var cardVal = [];
var cardIDs = [];
var cardBackFace = 0;
Array.prototype.cardMix = function()
{
var i = this.length, j, temp;
while (--i > 0)
{
j = Math.floor(Math.random() * (i + 1));
temp = this[j];
this[j] = this[i];
this[i] = temp;
}
}
function newBoard()
{
cardBackFace = 0;
var output = "";
cardArray.cardMix();
for (var i = 0; i < cardArray.length; i++)
{
output += '<div id="card_' + i + '" onclick="cardBackcard(this,\'' + cardArray[i].src + '\')"></div>';
}
document.getElementById('card_board').innerHTML = output;
}
function cardBackcard(tile, val) {
if (tile.innerHTML == "" && cardVal.length < 2) {
tile.style.background = '#FFF';
tile.innerHTML = val;
if (cardVal.length == 0) {
cardVal.push(val);
cardIDs.push(tile.id);
} else if (cardVal.length == 1) {
cardVal.push(val);
cardIDs.push(tile.id);
if (cardVal[0] == cardVal[1]) {
cardBackFace += 2;
cardVal = [];
cardIDs = [];
if (cardBackFace == cardArray.length) {
alert("Board cleared... generating new board");
document.getElementById('card_board').innerHTML = "";
newBoard();
}
} else {
function card2Back() {
var card_1 = document.getElementById(cardIDs[0]);
var card_2 = document.getElementById(cardIDs[1]);
card_1.style.background = 'url(cardQtion.jpg) no-repeat';
card_1.innerHTML = "";
card_2.style.background = 'url(cardQtion.jpg) no-repeat';
card_2.innerHTML = "";
cardVal = [];
cardIDs = [];
}
setTimeout(card2Back, 700);
}
}
}
}
</script>
</head>
<body>
<div id="card_board"></div>
<script>newBoard();</script>
</body>
</html>
benötigen Sie den Titel des Bildes in Komma zu setzen. Zum Beispiel hättest du 'background: url (" picture.jpg ");'. – Zachariel
@Zachariel Dies sind Anführungszeichen, keine Kommas, und sie sind in CSS nicht erforderlich. – Barmar
Es passiert wegen dieser Zeile: 'tile.innerHTML = val;' – Barmar