Question

寻找答案，并一直在试验但无济于事。我有一个下拉链接，在选择选项时自动打开弹出页面，但我无法弄清楚如何使弹出窗口居中。任何帮助表示赞赏：

<html>

<script language="JavaScript"> 
function goto(objSel) { 
if (objSel.selectedIndex > 0) { 
win = window.open(objSel.options[objSel.selectedIndex].value ,'','width=640, height=480, top=200, left=500, scrollbars=no, location=no, directories=no, status=no, menubar=no, toolbar=no, resizable=no, dependent=no');
win.focus();
}
}
//--> 
</script>

<form name="cityselect">
    <select name="menu" onchange="goto(this)" size="1">
        <option selected="selected">Select One</option>
        <option value="http://www.leeds.com">Leeds</option>
        <option value="http://www.manchester.com">Manchester</option>
    </select>
</form>

</html>

Answer 1

获取屏幕宽度和屏幕高度：

function goto(objSel) { 
    if (objSel.selectedIndex > 0) { 
        var topVar = Math.floor((screen.height - 480)/2);
        var leftVar = Math.floor((screen.width - 640)/2);
        var strOptions = "width=640, height=480, top=" + topVar + ", left=" + leftVar + ", scrollbars=no, location=no, directories=no, status=no, menubar=no, toolbar=no, resizable=no, dependent=no";
        win = window.open(objSel.options[objSel.selectedIndex].value ,'',strOptions);
        win.focus();
    }
}

使用onChange中心弹出窗口

1 个答案: