888starz 站點算法分析（待更新）

找到登陸包，看到需要兩個參數，comp_d 和 uLogin（下面password的加密算法相同。）

我先找的 uLogin，抓包定位，然後找到 crypt.encode 方法，跟進去，這個方法只會執行後面那個，所以從後面開始扒下。


直接給出算法

cb_utob = function(R) {
    let oe = null;
    return R.length < 2 ? (oe = R.charCodeAt(0),
    oe < 128 ? R : oe < 2048 ? fromCharCode(192 | oe >>> 6) + fromCharCode(128 | oe & 63) : fromCharCode(224 | oe >>> 12 & 15) + fromCharCode(128 | oe >>> 6 & 63) + fromCharCode(128 | oe & 63)) : (oe = 65536 + (R.charCodeAt(0) - 55296) * 1024 + (R.charCodeAt(1) - 56320),
    fromCharCode(240 | oe >>> 18 & 7) + fromCharCode(128 | oe >>> 12 & 63) + fromCharCode(128 | oe >>> 6 & 63) + fromCharCode(128 | oe & 63))
}
re_utob = /[\uD800-\uDBFF][\uDC00-\uDFFFF]|[^\x00-\x7F]/g

utob = function(R) {
    return R.replace(re_utob, cb_utob)
}


b64chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
cb_encode = function(R) {
    const oe = [0, 2, 1][R.length % 3]
      , le = R.charCodeAt(0) << 16 | (R.length > 1 ? R.charCodeAt(1) : 0) << 8 | (R.length > 2 ? R.charCodeAt(2) : 0);
    return [b64chars.charAt(le >>> 18), b64chars.charAt(le >>> 12 & 63), oe >= 2 ? "=" : b64chars.charAt(le >>> 6 & 63), oe >= 1 ? "=" : b64chars.charAt(le & 63)].join("")
}
btoa$1 = function(R) {
    return R.replace(/[\s\S]{1,3}/g, cb_encode)
}
_encode = function(R) {
    return btoa$1(utob(R))
}


console.log(_encode("[email protected]"))

comp_d參數好像很麻煩，找了蠻久，文章就到這裏，待更新吧