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是否有可能以某种方式配置 bundle 来为视网膜显示生成图像,比如@2x?

或者有人可以给我一个建议如何处理视网膜吗?

谢谢

4

1 回答 1

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根据nahakiole的评论,有两种解决方案:

您可以使用图片元素,该元素将提供为图像声明多个来源的语法。 http://w3c.github.io/html/semantics-embedded-content.html#the-picture-element

我们尝试过的另一种方法是,如果您可以保证图像存在,则使用retina.js 的修改版本,将_retina 添加到过滤器名称并检查具有此名称的图像是否存在。

nahakiole修改的视网膜.js版本:

/*-----------------------------------------------------------------------------------*/
/*	RETINA.JS
/*-----------------------------------------------------------------------------------*/
(function () {

    var regex = /(media\/cache\/filter_[A-Z]+)/i //Added this
    function t(e) {
        this.path = e;
        var t = this.path.split("."),
            n = t.slice(0, t.length - 1).join("."),
            r = t[t.length - 1];
        this.at_2x_path = (n + '.' + r).replace(regex, '$1_retina') //Changed that
    }
    function n(e) {
        this.el = e, this.path = new t(this.el.getAttribute("src"));
        var n = this;
        this.path.check_2x_variant(function (e) {
            e && n.swap()
        })
    }
    var e = typeof exports == "undefined" ? window : exports;
    e.RetinaImagePath = t, t.confirmed_paths = [], t.prototype.is_external = function () {
        return !!this.path.match(/^https?\:/i) && !this.path.match("//" + document.domain)
    }, t.prototype.check_2x_variant = function (e) {
        var n, r = this;
        if (this.is_external()) return e(!1);
        if (this.at_2x_path in t.confirmed_paths) return e(!0);
        n = new XMLHttpRequest, n.open("HEAD", this.at_2x_path), n.onreadystatechange = function () {
            return n.readyState != 4 ? e(!1) : n.status >= 200 && n.status <= 399 ? (t.confirmed_paths.push(r.at_2x_path), e(!0)) : e(!1)
        }, n.send()
    }, e.RetinaImage = n, n.prototype.swap = function (e) {
        function n() {
            t.el.complete ? (t.el.setAttribute("width", t.el.offsetWidth), t.el.setAttribute("height", t.el.offsetHeight), t.el.setAttribute("src", e)) : setTimeout(n, 5)
        }
        typeof e == "undefined" && (e = this.path.at_2x_path);
        var t = this;
        n()
    }, e.devicePixelRatio > 1 && (window.onload = function () {
        var e = document.getElementsByTagName("img"),
            t = [],
            r, i;
        for (r = 0; r < e.length; r++) i = e[r], t.push(new n(i))
    })
})();

于 2017-02-15T13:46:28.673 回答