onmousedown事件捕捉问题!如何捕捉窗口外鼠标的坐标.

yacus 2003-10-27 04:06:05

<HTML><HEAD><TITLE></TITLE>
<META http-equiv=Content-Type content="text/html; charset=gb2312">
<SCRIPT type=text/javascript>

s="再现";

function ss() {

if (window.event.button == 1) {
window.status = s+window.event.clientX;
}
}

document.onselectstart=function(){window.event.returnValue = false; window.event.cancelBubble = true;}
document.ondragstart = function(){return false}

</SCRIPT>

</HEAD>

<BODY scroll=no>
<IMG
onmousedown="this.onmousemove=function(){ss();};this.onmouseover=function(){s='在线'};this.onmouseout=function(){s='离线'}" onmouseup="this.onmousemove=null"
> </BODY></HTML>

////////////////////////////////////////////

为什么将上面的img改成div或别的元素,窗口外的事件就无法捕捉了呢?是不是该元素还有其它属性啊?

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Struggling 2003-10-27

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恭喜你，学习中·

leiaming 2003-10-27

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楼上的历害，支持！！

yacus 2003-10-27

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不知道就说不知道,别说无论如何都不能.多令人沮丧啊?

我搞定了..

setCapture();
//////////////
<HTML><HEAD><TITLE></TITLE>
<META http-equiv=Content-Type content="text/html; charset=gb2312">
<SCRIPT type=text/javascript>

s="再现";

function ss() {

if (window.event.button == 1) {
window.status = s+window.event.clientX;
}
}

</SCRIPT>

</HEAD>

<BODY>
<div
onmousedown="setCapture();this.onmousemove=function(){ss();};this.onmouseover=function(){s='在线'};this.onmouseout=function(){s='离线'}" onmouseup="alert(1);this.onmousemove=null"
unselectable="on">s</div></BODY></HTML>