使用jQuery获取所有元素CSS

时间:2014-01-08 21:46:19

标签: jquery css inheritance

我有一个简单的输入字段如下:

<input type="text" name="search_location" id="search_location" placeholder="Any Location" autocomplete="off">

使用jQuery,我试图让所有样式与它相关联。这个代码看起来像这样(从stackoverflow thread获得):

// Get all styles associated with element
var style = css(jQuery("#elementToGetAllCSS"));

function css(a) {
    var sheets = document.styleSheets, o = {};
    for (var i in sheets) {
        var rules = sheets[i].rules || sheets[i].cssRules;
        for (var r in rules) {
            if (a.is(rules[r].selectorText)) {
                o = $.extend(o, css2json(rules[r].style), css2json(a.attr('style')));
            }
        }
    }
    return o;
}

function css2json(css) {
    var s = {};
    if (!css) return s;
    if (css instanceof CSSStyleDeclaration) {
        for (var i in css) {
            if ((css[i]).toLowerCase) {
                s[(css[i]).toLowerCase()] = (css[css[i]]);
            }
        }
    } else if (typeof css == "string") {
        css = css.split("; ");
        for (var i in css) {
            var l = css[i].split(": ");
            s[l[0].toLowerCase()] = (l[1]);
        }
    }
    return s;
}

// Apply styles to target element
var style = css(jQuery("#elementToGetAllCSS"));
jQuery("#search_location").css(style);

一切都很好,除非涉及CSS伪选择器,例如:before和:之后我得到一个“无法识别的表达式”控制台错误。

有什么想法吗?

编辑 - 我的解决方案

下面的答案是有效的,尽管我有点担心未来的答案如何。考虑到这一点,有一种似乎工作正常的替代方案:

$.fn.getStyleObject = function(){
        var dom = this.get(0);
        var style;
        var returns = {};
        if(window.getComputedStyle){
            var camelize = function(a,b){
                return b.toUpperCase();
            };
            style = window.getComputedStyle(dom, null);
            for(var i = 0, l = style.length; i < l; i++){
                var prop = style[i];
                var camel = prop.replace(/\-([a-z])/g, camelize);
                var val = style.getPropertyValue(prop);
                returns[camel] = val;
            };
            return returns;
        };
        if(style = dom.currentStyle){
            for(var prop in style){
                returns[prop] = style[prop];
            };
            return returns;
        };
        return this.css();
    }

var styles = $('elementToGetAllCSS').getStyleObject();
  this.css('search_location);

1 个答案:

答案 0 :(得分:2)

您可以通过检查选择器是否包含::

来过滤掉循环中的伪选择器
for (var r in rules) {
    if (typeof rules[r] == 'object' && 
        'selectorText' in rules[r] && 
        rules[r].selectorText.indexOf('::') == -1 &&
        rules[r].selectorText.indexOf(':hover') == -1
    ) {
        if (a.is(rules[r].selectorText)) {
           o = $.extend(o, css2json(rules[r].style), css2json(a.attr('style')));
        }
    }
}

FIDDLE