为什么字符在 Internet Explorer 中“未定义”，而在其他浏览器中却没有
[英]Why characters are 'undefined' in internet explorer but not in other browsers
问题描述
Please take a look at the site tdsoft.se The script on that page works in opera, firefox chrome etc and prints out "random_1" as it is suposed to do, but in internet explorer it just prints out (" undefinedundefinedundefinedundefinedundefinedundefinedundefinedundefined "), that is 'undefined' for each letter.
请查看站点tdsoft.se 该页面上的脚本适用于歌剧、firefox chrome 等，并按预期打印出“random_1”，但在 Internet Explorer 中它只是打印出（“ undefinedundefinedundefinedundefinedundefinedundefinedundefinedundefinedundefined ”），即对于每个字母都是“未定义的”。 My question is if some of you bright fellows out there might know the answer to this problem?我的问题是，你们中的一些聪明人是否知道这个问题的答案？
EDIT_ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ ____编辑_ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ ____
Here's the code这是代码
<html>
<head>
<script language="javascript" type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
var txt;
var buildName = "";
var xmlhttp;
/**
* Find a longest common subsenquence.
*
* Note: this is not necessarily the only possible longest common subsequence though!
*/
function lcs(listX, listY) {
return lcsBackTrack(
lcsLengths(listX, listY),
listX, listY,
listX.length, listY.length);
}
/**
* Iteratively memoize a matrix of longest common subsequence lengths.
*/
function lcsLengths(listX, listY) {
var lenX = listX.length;
var lenY = listY.length;
// Initialize a lenX+1 x lenY+1 matrix
var memo = [lenX+1];
for (var i = 0; i < lenX+1; i++) {
memo[i] = [lenY+1];
for (var j = 0; j < lenY+1; j++) {
memo[i][j] = 0;
}
}
// Memoize the lcs length at each position in the matrix
for (var i = 1; i < lenX+1; i++) {
for (var j = 1; j < lenY+1; j++) {
if (listX[i-1] == listY[j-1]) {
memo[i][j] = memo[i-1][j-1] + 1;
}
else {
memo[i][j] = Math.max(
memo[i][j-1],
memo[i-1][j]);
}
}
}
return memo;
}
/**
* Recursively read back a memoized matrix of longest common subsequence lengths
* to find a longest common subsequence.
*/
function lcsBackTrack(memo, listX, listY, posX, posY) {
// base case
if (posX == 0 || posY == 0) {
return "";
}
// matcth => go up and left
else if (listX [posX-1] == listY[posY-1]) {
return lcsBackTrack(memo, listX, listY, posX-1, posY-1) + listX[posX-1];
}
else {
// go up
if (memo[posX][posY-1] > memo[posX-1][posY]) {
return lcsBackTrack(memo, listX, listY, posX, posY-1);
}
// go left
else {
return lcsBackTrack(memo, listX, listY, posX-1, posY);
}
}
}
function loadXMLDoc(url,cfunc)
{
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=cfunc;
xmlhttp.open("GET",url,true);
xmlhttp.send();
}
function myFunction()
{
loadXMLDoc("http://tdsoft.se/testni.html",handleXML);
}
var checkState = function(xmlhttp, callback) {
try{
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
callback();
}
else {
// Check back again 1 sec later
setTimeout(checkState, 1000);
}
}
catch(err){
setTimeout(checkState, 1000);
}
};
function handleXML()
{
checkState(xmlhttp, function() {
txt=xmlhttp.responseText;
buildName = "random_1";
var myvar = "";
txt = "" + txt;
var lcsString = lcs(txt, buildName);
document.write(lcsString);
});
}
</script>
</head>
<body onLoad="myFunction()">
</body>
</html>
3 个解决方案
解决方案1
1 2011-07-11 15:46:13
[lenY + 1] , for one, does not initialize an array with lenY + 1 elements. [lenY + 1]一方面，不初始化具有lenY + 1元素的数组。 It initializes an array with one element set to lenY + 1 .它初始化一个数组，其中一个元素设置为lenY + 1 。 Not that that matters, because you set them anyways, to zero... just change that to [] both times.没关系，因为无论如何你都将它们设置为零......只需将其更改为[]两次。
I'm having trouble figuring out your code, but I believe the problem is that IE only allows you to access string characters using charAt , and not using the bracket notation, which you appear to be using here:我在弄清楚您的代码时遇到了麻烦，但我认为问题在于 IE 只允许您使用charAt访问字符串字符，而不是使用括号表示法，您似乎在这里使用它：
listX[i-1] == listY[j-1]
and here:和这里：
else if (listX [posX-1] == listY[posY-1]) {
So those comparisons would always return true .所以这些比较总是会返回true 。 Could that be the issue?这可能是问题吗？
解决方案2
1 已采纳 2011-07-11 15:57:44
This is not working in IE, listX [posX-1] .这在 IE, listX [posX-1]中不起作用。 The result of this is "undefined" so you sould use another way to get the char like chatAt() method结果是“未定义”，因此您应该使用另一种方法来获取类似chatAt()方法的字符
解决方案3
-1 2012-01-25 19:51:46
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