Question

在Javascript函数中，我有两个结构相似的条件。我想知道你是否可以将它们组合成一个。代码来自我的随机加密安全数字/字符生成器。这是Java脚本代码的片段：

function randomString(length, charset, allowRandomSourceFallback) {
    var i,
        result = "";

    // First we're going to try to use a built-in CSPRNG
    if (window.crypto && window.crypto.getRandomValues) {
        values = new Uint32Array(length);
        window.crypto.getRandomValues(values);

        for (i = 0; i < length; i++) {
            result += charset[values[i] % charset.length];
        }
    }
    // IE calls it msCrypto (built-in CSPRNG)
    else if (window.msCrypto && window.msCrypto.getRandomValues) {
        values = new Uint32Array(length);
        window.msCrypto.getRandomValues(values);

        for (i = 0; i < length; i++) {
            result += charset[values[i] % charset.length];
        }
    }
    // -> No built-in functionality -> use the function Math.random()
    else {
        for (i = 0; i < length; i++) {
            result += charset[Math.floor(Math.random() * charset.length)];
        }
    }
    return result;
}

Answer 1

为什么不检查可用的内容然后使用它：

var availableCrypto = window.crypto || window.msCrypto; // will get whatever is available

然后像这样使用它：

availableCrypto.getRandomValues(values);

你走了：

function randomString(length, charset, allowRandomSourceFallback) {
    var i,
        result = "",
        availableCrypto = window.crypto || window.msCrypto;

    // First we're going to try to use a built-in CSPRNG
    if (availableCrypto && availableCrypto.getRandomValues) {
        values = new Uint32Array(length);
        availableCrypto.getRandomValues(values);

        for (i = 0; i < length; i++) {
            result += charset[values[i] % charset.length];
        }
    }
    // -> No built-in functionality -> use the function Math.random()
    else {
        for (i = 0; i < length; i++) {
            result += charset[Math.floor(Math.random() * charset.length)];
        }
    }
    return result;
}

Answer 2

而不是将它们组合在一起，只声明一个加密函数，如下面的

var crypto = null;
if (window.crypto && window.crypto.getRandomValues) {
        crypto = window.crypto;        
    }
    // IE calls it msCrypto (built-in CSPRNG)
    else if (window.msCrypto && window.msCrypto.getRandomValues) {
        crypto = window.msCrypto ;
    }

if(crypto != null){
        values = new Uint32Array(length);
        crypto.getRandomValues(values);

        for (i = 0; i < length; i++) {
            result += charset[values[i] % charset.length];
        }
}
    else {
        for (i = 0; i < length; i++) {
            result += charset[Math.floor(Math.random() * charset.length)];
        }
    }
    return result;

两个相似的if条件合二为一

2 个答案: