使用javascript脚本定位CSS弹出窗口

这是一个什么样的小提琴我想我的弹出窗口看起来像。。。这很有效：
（两个

显示

都设置为：

block；

），并且显示效果完美

当我在代码中运行它时，我使用一个JS脚本来弹出窗口

我的HTML：

  <a href="#" onclick="popup('popUpDiv')">Click to Open CSS Pop Up</a> 

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<!--POPUP-->    
<script type="text/javascript" src="css-pop.js"></script>

<div id="blanket" style="display:none;"></div>

<div id="popUpDiv" style="display:none;">
<a href="#" onclick="popup('popUpDiv')" class="popclose">Click to Close CSS Pop Up</a>
</div>  
<!-- END POPUP -->

下面是JS代码：

function toggle(div_id) {
    var el = document.getElementById(div_id);
    if ( el.style.display == 'none' ) { el.style.display = 'block';}
    else {el.style.display = 'none';}
}
function blanket_size(popUpDivVar) {
    if (typeof window.innerWidth != 'undefined') {
        viewportheight = window.innerHeight;
    } else {
        viewportheight = document.documentElement.clientHeight;
    }
    if ((viewportheight > document.body.parentNode.scrollHeight) && (viewportheight > document.body.parentNode.clientHeight)) {
        blanket_height = viewportheight;
    } else {
        if (document.body.parentNode.clientHeight > document.body.parentNode.scrollHeight) {
            blanket_height = document.body.parentNode.clientHeight;
        } else {
            blanket_height = document.body.parentNode.scrollHeight;
        }
    }
    var blanket = document.getElementById('blanket');
    blanket.style.height = blanket_height + 'px';
    var popUpDiv = document.getElementById(popUpDivVar);
    popUpDiv_height=blanket_height/2-200;//200 is half popup's height
    popUpDiv.style.top = popUpDiv_height + 'px';
}
function window_pos(popUpDivVar) {
    if (typeof window.innerWidth != 'undefined') {
        viewportwidth = window.innerHeight;
    } else {
        viewportwidth = document.documentElement.clientHeight;
    }
    if ((viewportwidth > document.body.parentNode.scrollWidth) && (viewportwidth > document.body.parentNode.clientWidth)) {
        window_width = viewportwidth;
    } else {
        if (document.body.parentNode.clientWidth > document.body.parentNode.scrollWidth) {
            window_width = document.body.parentNode.clientWidth;
        } else {
            window_width = document.body.parentNode.scrollWidth;
        }
    }
    var popUpDiv = document.getElementById(popUpDivVar);
    window_width=window_width/2-200;//200 is half popup's width
    popUpDiv.style.left = window_width + 'px';
}
function popup(windowname) {
    blanket_size(windowname);
    window_pos(windowname);
    toggle('blanket');
    toggle(windowname);     
}

我对JS代码的理解是

#毯子

和

#popUpDiv

更改

显示：无至<代码>显示：块。。。但是弹出窗口没有像在小提琴上那样放在正确的位置？我不知道为什么。。。JS中一定有什么东西导致了这个。。。因为HTML和CSS正确地定位了弹出窗口。。。有什么想法吗

谢谢
哦，我的。。。首先，您应该将其从css中删除：

margin-left: -200px;
margin-top: -200px;

然后替换为：

popUpDiv_height=blanket_height/2-200;//200 is half popup's height

因此：

popUpDiv_height=viewportheight/2-200;//200 is half popup's height

因为这是一页的全高。您只需要视口的高度

popUpDiv_height=viewportheight/2-200;//200 is half popup's height