javascript - jQuery是否可以获取与元素关联的所有CSS样式？(Can jQuery get all CSS styles associated with an element?)

Question

Is there a way in jQuery to get all CSS from an existing element and apply it to another without listing them all?

(jQuery中是否有一种方法可以从现有元素中获取所有CSS并将其应用于另一个元素而不列出所有元素？)

I know it would work if they were a style attribute with attr() , but all of my styles are in an external style sheet.

(我知道如果它们是attr()的样式属性，它将起作用，但是我所有的样式都在外部样式表中。)

  ask by alex translate from so

Answer 1

A couple years late, but here is a solution that retrieves both inline styling and external styling:

(迟了几年，但是这里有一个可以同时检索内联样式和外部样式的解决方案：)

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;
}

Pass a jQuery object into css() and it will return an object, which you can then plug back into jQuery's $().css() , ex:

(将jQuery对象传递到css() ，它将返回一个对象，然后您可以将其插入jQuery的$().css() ，例如：)

var style = css($("#elementToGetAllCSS"));
$("#elementToPutStyleInto").css(style);

:)

(:))