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我一直在處理這段代碼一段時間,出於某種原因,我無法告訴它當我按下更改圖像按鈕時不會更改顯示的圖像。誰能幫忙?使用Javascript來更改網頁上的圖像
這是JavaScript文件
var imgArray = new Array();
imgArray[0] = new Image();
imgArray[0].src = "images/img/red.gif";
imgArray[1] = new Image();
imgArray[1].src = "images/img/red_and_yellow.gif";
imgArray[2] = new Image();
imgArray[2].src = "images/img/green.gif";
imgArray[3] = new Image();
imgArray[3].src = "images/img/yellow.gif";
imgArray[4] = new Image();
imgArray[4].src = "images/img/red.gif";
/*------------------------------------*/
function nextImage(element) {
var img = document.getElementById(element);
for (var i = 0; i < imgArray.length; i++) {
if (imgArray[i].src == img.src) // << check this
{
if (i === imgArray.length) {
document.getElementById(element).src = imgArray[0].src;
break;
}
document.getElementById(element).src = imgArray[i + 1].src;
break;
}
}
}
<!doctype html>
<html>
<body>
<h1> Press the button to change the lights. </h1>
<div id="lights">
<img src="red.gif" alt="" id="mainImg">
</div>
<div id="imglist">
<a href="javascript:nextImage('mainImg')">
<img src="next_img.png" alt="">
</a>
</body>
</html>
沒有意義,使用for循環比較之前。如果你知道起始索引只是增加一個計數器。不需要所有這些檢查。 – epascarello