- 一个拖动拼图游戏的代码小问题想不通
-
html代码：
<!doctype html>
Sliding Puzzle
<br> .picture {<br> border: 1px solid black;<br> }<br>Sliding Puzzle
Easy Hard
js代码：var context = document.getElementById('puzzle').getContext('2d');var img = new Image();
img.src = './img/dimetrodon.jpg';
img.addEventListener('load', drawTiles, false);var boardSize = document.getElementById('puzzle').width;
var tileCount = document.getElementById('scale').value;var tileSize = boardSize / tileCount;
var clickLoc = new Object;
clickLoc.x = 0;
clickLoc.y = 0;var emptyLoc = new Object;
emptyLoc.x = 0;
emptyLoc.y = 0;var solved = false;
var boardParts = new Object;
setBoard();document.getElementById('scale').onchange = function() {
tileCount = this.value;
tileSize = boardSize / tileCount;
setBoard();
drawTiles();
};document.getElementById('puzzle').onmousemove = function(e) {
clickLoc.x = Math.floor((e.pageX - this.offsetLeft) / tileSize);
clickLoc.y = Math.floor((e.pageY - this.offsetTop) / tileSize);
};document.getElementById('puzzle').onclick = function() {
if (distance(clickLoc.x, clickLoc.y, emptyLoc.x, emptyLoc.y) == 1) {
slideTile(emptyLoc, clickLoc);
drawTiles();
}
if (solved) {
setTimeout(function() {alert("You solved it!");}, 500);
}
};function setBoard() {
boardParts = new Array(tileCount);
for (var i = 0; i < tileCount; ++i) {
boardParts[i] = new Array(tileCount);
for (var j = 0; j < tileCount; ++j) {
boardParts[i][j] = new Object;
boardParts[i][j].x = (tileCount - 1) - i;
boardParts[i][j].y = (tileCount - 1) - j;
}
}
emptyLoc.x = boardParts[tileCount - 1][tileCount - 1].x;
emptyLoc.y = boardParts[tileCount - 1][tileCount - 1].y;
solved = false;
}function drawTiles() {
context.clearRect ( 0 , 0 , boardSize , boardSize );
for (var i = 0; i < tileCount; ++i) {
for (var j = 0; j < tileCount; ++j) {
var x = boardParts[i][j].x;
var y = boardParts[i][j].y;
if(i != emptyLoc.x || j != emptyLoc.y || solved == true) {
context.drawImage(img, x * tileSize, y * tileSize, tileSize, tileSize,
i * tileSize, j * tileSize, tileSize, tileSize);
}
}
}
}function distance(x1, y1, x2, y2) {
return Math.abs(x1 - x2) + Math.abs(y1 - y2);
}function slideTile(toLoc, fromLoc) {
if (!solved) {
boardParts[toLoc.x][toLoc.y].x = boardParts[fromLoc.x][fromLoc.y].x;
boardParts[toLoc.x][toLoc.y].y = boardParts[fromLoc.x][fromLoc.y].y;
boardParts[fromLoc.x][fromLoc.y].x = tileCount - 1;
boardParts[fromLoc.x][fromLoc.y].y = tileCount - 1;
// var a = toLoc.x;
// var b = toLoc.y;
toLoc.x = fromLoc.x;
toLoc.y = fromLoc.y;
// fromLoc.x = a;
// fromLoc.y = b;//这边为什么不用给emptyLoc改变坐标？
checkSolved();
}
}function checkSolved() {
var flag = true;
for (var i = 0; i < tileCount; ++i) {
for (var j = 0; j < tileCount; ++j) {
if (boardParts[i][j].x != i || boardParts[i][j].y != j) {
flag = false;
}
}
}
solved = flag;
}
请问为什么js中的emptyLoc的x和y不需要更新，或者他的x和y是怎么更新的，他代码里面emptyLoc.x和emptyLoc.y的值貌似都是2，没有变化，不更新的话没办法继续判断距离distance==1啊，为什么它程序可以正确运行，还是我理解错了？
slideTile(emptyLoc, clickLoc);在这里传入的清空位置和点击位置
而 clickLoc 在你鼠标移动到某个puzzle上面，就根据它的位置确定了。
document.getElementById('puzzle').onmousemove = function(e) {
clickLoc.x = Math.floor((e.pageX - this.offsetLeft) / tileSize);
clickLoc.y = Math.floor((e.pageY - this.offsetTop) / tileSize);
};
-
- iron_nerve clickLoc的位置了，谢谢啦~
- 一年多之前 回复
-
- iron_nerve 谢谢大佬~我想通了，我理解反了，其实那个位置信息只需要更新emptyLoc的，而他toLoc就是更新的enptyLoc的位置，我理解成更新的
- 一年多之前 回复
-
- iron_nerve 哦，您的意思是，在方法内fromLoc的位置信息也会跟着onmousemove而变化是吧，谢谢
- 一年多之前 回复
-
- iron_nerve // var a = toLoc.x; // var b = toLoc.y; toLoc.x = fromLoc.x; toLoc.y = fromLoc.y; // fromLoc.x = a; // fromLoc.y = b;//这边为什么不用给emptyLoc改变坐标？ 他这边只是给toLoc.x = fromLoc.x; toLoc.y = fromLoc.y;，这样更新，而没有给fromLoc也就是emptyLoc的x,y更新位置信息，emptyLoc的位置信息好像一直都是emptyLoc.x=0，emptyLoc.y=0, 请问您他的信息是在哪儿更新的？
- 一年多之前 回复
// var a = toLoc.x;
// var b = toLoc.y;
toLoc.x = fromLoc.x;
toLoc.y = fromLoc.y;
// fromLoc.x = a;
// fromLoc.y = b;//这边为什么不用给emptyLoc改变坐标？
他这边只是给toLoc.x = fromLoc.x;
toLoc.y = fromLoc.y;，这样更新，而没有给fromLoc也就是emptyLoc的x,y更新位置信息，emptyLoc的位置信息好像一直都是emptyLoc.x=0，emptyLoc.y=0,
请问您他的信息是在哪儿更新的？