javascript - 使用 JavaScript 脚本定位 CSS 弹出窗口

Question

这是我希望我的弹出窗口看起来像的 FIDDLE...这很好用: http://jsfiddle.net/mib92/x6yev5a0/
(display 都设置为:block; )并且显示完美。

当我在我的代码中运行它时，我使用 JS 脚本来执行弹出窗口。

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 -->

我的 CSS:(与没有 display: block; 的 FIDDLE 相同，因为它在 HTML 中设置为 display:none;)

#blanket {
background-color:#ff0024;
opacity: 0.65;
*background:none;
position:fixed;
z-index: 9001;
top:0px;
left:0px;
width:100%;
height:100%;
}

#popUpDiv {
 position:fixed;
 background-color:#ccc;
 width:400px;
 height:400px;
 border:5px solid #000;
 z-index: 9002;
 top:50%;
 left:50%;
 margin-left: -200px;
 margin-top: -200px;    
}

这是 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#blanket和#popUpDiv将display:none;改为display :block; ...但是弹出窗口没有像在 Fiddle 中那样定位在正确的位置？我不知道为什么......一定是 JS 中有什么东西导致了这个......因为 HTML 和 CSS 正确定位了弹出窗口......有什么想法吗？

谢谢

Answer 1

哦，我的...首先，你应该从 css 中删除它:

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

然后替换它:

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

有了那个:

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

因为 blanket_height - 它是页面的全高。你只需要视口(viewport)的高度