javascript 在随机位置定位图像
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Positioning images in random places
提问by user2484181
I want to position random images in random places of the screen. I tried this:
我想在屏幕的随机位置放置随机图像。我试过这个：
$(document).ready(showLetter);
var imgsArray = ["A1", "C1", "F1", "J1", "K1", "L1", "S1", "?1"];
function generateRandomForArray() {
var num = Math.floor(Math.random() * 8);
return num;
}
function generateRandom() {
var num = Math.floor(Math.random() * 400);
return num;
}
function showLetter() {
var letter = imgsArray[generateRandomForArray()];
$("div").append("<img src='imgs/" + letter + ".png'>");
var left = generateRandom();
var top = generateRandom();
$("div").last().css({"top": top + "px", "left": left + "px"});
}
The images appear on top left of the screen. Once again: I want to position them in random places of the screen. I don't know much about CSS. Any ideas? Thanks in advance!
图像出现在屏幕的左上方。再一次：我想将它们放置在屏幕的随机位置。我对 CSS 了解不多。有任何想法吗？提前致谢！
回答by Vinay
Sounds like positioning issue for me. Try like this, i.e., I added position:absolute
对我来说听起来像是定位问题。像这样尝试，即，我添加了position:absolute
function showLetter() {
var letter = imgsArray[generateRandomForArray()];
$("div").append("<img src='GameHTML5/images/" + letter + ".png'>");
var left = generateRandom();
var top = generateRandom();
$("div").last().css({"position":"absolute","top": top + "px", "left": left + "px"});
}
回答by user2484181
function generateRandomForArray(min, max) {
var num = Math.random() * (max - min) + min;
return Math.floor(num);
}
//generateRandomForArray(0, 8);
//3
//generateRandomForArray(0, 8);
//0
//generateRandomForArray(0, 8);
//2
//generateRandomForArray(0, 8);
//4
//generateRandomForArray(0, 8);
//5
//generateRandomForArray(0, 8);
//7
//generateRandomForArray(0, 8);
//4
