我正在设置一个等于图像的变量,然后在函数中调用它。 这不起作用,因为根据javascript错误控制台,它不会将图像变量识别为null
var image = document.getElementById("gc");
var duration = 3000; /* fade duration in millisecond */
var hidtime = 2000; /* time to stay hidden */
var showtime = 2000; /* time to stay visible */
var image_tracker = 1
function change() {
if (image_tracker == 1) {
image.src = "IMG_1267.jpg"
image_tracker = 2
} else if (image_tracker == 2) {
image.src = "IMG_1239.jpg"
image_tracker = 3
} else {
image.src = "IMG_1263.jpg"
image_tracker = 1
}
}
然而这很好用
var duration = 3000; /* fade duration in millisecond */
var hidtime = 2000; /* time to stay hidden */
var showtime = 2000; /* time to stay visible */
var image_tracker = 1
function change() {
var image = document.getElementById("gc");
if (image_tracker == 1) {
image.src = "IMG_1267.jpg"
image_tracker = 2
} else if (image_tracker == 2) {
image.src = "IMG_1239.jpg"
image_tracker = 3
} else {
image.src = "IMG_1263.jpg"
image_tracker = 1
}
}
var timer = setInterval('change()', 3000);
答案 0 :(得分:6)
在您的代码中,即使在DOM完全加载之前,您的JavaScript也会立即加载。当它尝试搜索#gc
时,它无法找到它,因为它没有加载到DOM中。
解决方案是在页面完全加载时设置image
变量。离。
var image;
window.onload = function() {
image = document.getElementById("gc");
}
function change() {
....