在javascript/jquery IE与Chrome和FF中读取@font-face css

我试图读入一个javascript变量，即包含@font-face的（本地）样式表的CSS。 我设法让它在FF en Chrome中工作，但在IE或Safari中却不行

这是我的例程（另请参见--在不同的浏览器中尝试）

函数getFonts（obj，doc）{ //列出所有@font-face项的cssText。 var o=obj |{}， sheets=doc.styleSheets， i=板材长度，j，规则；

---

虽然（0颠倒检查

工作表[i]。规则和
工作表[i]。cssRules
-IE两者都匹配，但
CSSFONFTACHERULE
仅在后者中找到（在IE9+中测试）


函数getFonts（obj，doc）{
//Maakt een lijst遇到了css van alle@font-face项目。
var o=obj |{}，
sheets=doc.styleSheets，
规则=null，
i=板材长度，j；
你可以使用BrowserStack.com在其他操作系统上进行测试。先生，你是我今天的英雄。我知道这可能很简单，但这很简单……哈哈！
function getFonts (obj, doc) {
    // Makes a list of the cssText of all @font-face items.
    var o = obj || {},
    sheets = doc.styleSheets,
    i = sheets.length, j, rules;

    while( 0 <= --i ){
        rules = sheets[i].rules || sheets[i].cssRules || [];
        j = rules.length;

        while( 0 <= --j ){
            // Alternatives for the if below:
            // - (worked only in Chrome): if ( rule[j].constructor.name === 'CSSFontFaceRule' )
            // - if ( rule[j].style.cssText && rule[j].style.cssText.match('src:') )
            // IE just gives NO [object CSSFontFaceRule] !!!
            if( rules[j].toString() == "[object CSSFontFaceRule]" ) {
                // rule.style.fontFamily works in Chrome chrome; Not in FF. For IE I don't know yet.
                fontFamily = rules[j].style.fontFamily || rules[j].style.cssText.match(/font-family\s*:\s*([^;\}]*)\s*[;}]/i)[1]

                // To prevent duplicates we use the cssText as key 
                o[ fontFamily ] = o[ fontFamily ] || {} ;
                o[ fontFamily ][rules[j].style.cssText] = rules[j].style.cssText ;
            }
        }
    }
    return o;
}

function getFonts (obj, doc) {
    // Maakt een lijst met de css van alle @font-face items.
    var o = obj || {},
    sheets = doc.styleSheets,
    rule = null,
    i = sheets.length, j;

    while( 0 <= --i ){
        rules = sheets[i].cssRules || sheets[i].rules || [];
        j = rules.length;

        while( 0 <= --j ){
            // Alternatives for the if below:
            // - if ( rule[j].style.cssText && rule[j].style.cssText.match('src:') )
            // - (worked only in Chrome): if ( rule[j].constructor.name === 'CSSFontFaceRule' )
            // IE just gives NO [object CSSFontFaceRule] !!!
            if( rules[j].toString() == "[object CSSFontFaceRule]" ) {
                // rule.style.fontFamily works in Chrome chrome; Not in FF. For IE I don't know yet.
                fontFamily = rules[j].style.fontFamily || rules[j].style.cssText.match(/font-family\s*:\s*([^;\}]*)\s*[;}]/i)[1]

                // To prevent duplicates we use the cssText as key 
                o[ fontFamily ] = o[ fontFamily ] || {} ;
                o[ fontFamily ][rules[j].style.cssText] = rules[j].style.cssText ;
            }
        }
    }
    return o;
}

console.log(getFonts(null, document));