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我在寫jQuery代碼。但我現在有一個問題。當我編寫這些代碼時,我想關閉彈出窗口的可調整大小的選項。但事實並非如此。我能做什麼 ?Jquery popup可調整大小的參數
Jquery的代碼
<script type="text/javascript" language="javascript">
<!--
function popupfs(url)
{
alert($(window).width());
params = 'top=0, left=0'
if($(window).width() < 1200){
params += ', width='+(962)+', height='+(656);
}else if($(window).width() > 1200 && $(window).width() < 1550){
params += ', width='+(1100)+', height='+(750);
}else if($(window).width() > 1550){
params += ', width='+(1547)+', height='+(1055);
}
params += ', directories=no';
params += ', location=no';
params += ', menubar=no';
params += ', resizable=no';
newwin=window.open(url,'windowname4', params);
if (window.focus) {newwin.focus()}
return false;
}// -->
</script>
HTML代碼
<body>
<div id="external" >
</div>
<a href="javascript: void(0)" onclick="popupfs('2/index.html')">Ekranın Ortasına Popup Açma</a>
</body>