经过分析，我们发现这里有两个参数发生变化，经过推断，这里的参数可能是使用js生成的，所以我们这里要对接口进行分析，破解，得到这里的值
注意： 在js逆向中，不是你要精通js才可以js逆向，你要懂一点js就可以做逆向，重要的逆向思维，对问题的思考方式
Copy function a(r) {
var t = r.match(/[\uD800-\uDBFF][\uDC00-\uDFFF]/g);
if (null === t) {
var a = r.length;
a > 30 && (r = "" + r.substr(0, 10) + r.substr(Math.floor(a / 2) - 5, 10) + r.substr(-10, 10))
} else {
for (var C = r.split(/[\uD800-\uDBFF][\uDC00-\uDFFF]/), h = 0, f = C.length, u = []; f > h; h++)
"" !== C[h] && u.push.apply(u, e(C[h].split(""))),
h !== f - 1 && u.push(t[h]);
var g = u.length;
g > 30 && (r = u.slice(0, 10).join("") + u.slice(Math.floor(g / 2) - 5, Math.floor(g / 2) + 5).join("") + u.slice(-10).join(""))
}
var l = void 0
, d = "" + String.fromCharCode(103) + String.fromCharCode(116) + String.fromCharCode(107);
l = null !== i ? i : (i = o.common[d] || "") || "";
for (var m = l.split("."), S = Number(m[0]) || 0, s = Number(m[1]) || 0, c = [], v = 0, F = 0; F < r.length; F++) {
var p = r.charCodeAt(F);
128 > p ? c[v++] = p : (2048 > p ? c[v++] = p >> 6 | 192 : (55296 === (64512 & p) && F + 1 < r.length && 56320 === (64512 & r.charCodeAt(F + 1)) ? (p = 65536 + ((1023 & p) << 10) + (1023 & r.charCodeAt(++F)),
c[v++] = p >> 18 | 240,
c[v++] = p >> 12 & 63 | 128) : c[v++] = p >> 12 | 224,
c[v++] = p >> 6 & 63 | 128),
c[v++] = 63 & p | 128)
}
for (var w = S, A = "" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(97) + ("" + String.fromCharCode(94) + String.fromCharCode(43) + String.fromCharCode(54)), b = "" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(51) + ("" + String.fromCharCode(94) + String.fromCharCode(43) + String.fromCharCode(98)) + ("" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(102)), D = 0; D < c.length; D++)
w += c[D],
w = n(w, A);
return w = n(w, b),
w ^= s,
0 > w && (w = (2147483647 & w) + 2147483648),
w %= 1e6,
w.toString() + "." + (w ^ S)
}
Copy import requests
import js2py
context = js2py.EvalJs()
class BaiDuTranslater(object):
"""
百度翻译爬虫
"""
def __init__(self, query):
# 初始化
self.url = "https://fanyi.baidu.com/basetrans"
self.query = query
self.headers = {
"User-Agent": "Mozilla/5.0 (iPhone; CPU iPhone OS 11_0 like Mac OS X) AppleWebKit/604.1.38 (KHTML, like Gecko) Version/11.0 Mobile/15A372 Safari/604.1",
"Referer": "https://fanyi.baidu.com/",
"Cookie": "BAIDUID=714BFAAF02DA927F583935C7A354949A:FG=1; BIDUPSID=714BFAAF02DA927F583935C7A354949A; PSTM=1553390486; delPer=0; PSINO=5; H_PS_PSSID=28742_1463_21125_18559_28723_28557_28697_28585_28640_28604_28626_22160; locale=zh; from_lang_often=%5B%7B%22value%22%3A%22en%22%2C%22text%22%3A%22%u82F1%u8BED%22%7D%2C%7B%22value%22%3A%22zh%22%2C%22text%22%3A%22%u4E2D%u6587%22%7D%5D; to_lang_often=%5B%7B%22value%22%3A%22en%22%2C%22text%22%3A%22%u82F1%u8BED%22%7D%2C%7B%22value%22%3A%22zh%22%2C%22text%22%3A%22%u4E2D%u6587%22%7D%5D; REALTIME_TRANS_SWITCH=1; FANYI_WORD_SWITCH=1; HISTORY_SWITCH=1; SOUND_SPD_SWITCH=1; SOUND_PREFER_SWITCH=1; Hm_lvt_afd111fa62852d1f37001d1f980b6800=1553658863,1553766321,1553769980,1553770442; Hm_lpvt_afd111fa62852d1f37001d1f980b6800=1553770442; Hm_lvt_64ecd82404c51e03dc91cb9e8c025574=1553766258,1553766321,1553769980,1553770442; Hm_lpvt_64ecd82404c51e03dc91cb9e8c025574=1553770442"
}
def make_sign(self):
# js逆向获取sign的值
# 读取js文件
with open("translate.js", "r", encoding="utf-8") as f:
# 添加至上下文
context.execute(f.read())
# 调用js中的函数生成sign
sign = context.a(self.query)
# 将sign加入到data中
return sign
def make_data(self, sign):
data = {
"query": self.query,
"from": "en",
"to": "zh",
"token": "6f5c83b84d69ad3633abdf18abcb030d",
"sign": sign
}
return data
def get_content(self, data):
# 发送请求获取响应
response = requests.post(
url=self.url,
headers=self.headers,
data=data
)
return response.json()["trans"][0]["dst"]
def run(self):
"""运行程序"""
# 获取sign的值
sign = self.make_sign()
# 构建参数
data = self.make_data(sign)
# 获取翻译内容
content = self.get_content(data)
print(content)
if __name__ == '__main__':
query = input("请输入您要翻译的内容:")
translater = BaiDuTranslater(query)
translater.run() 解决办法：回到浏览器，我们查找 r 的值,并将生成 r 的值的函数，加入到我们之前创建的 js 文件中，放在 G 函数的上面
Copy function n(r, o) {
for (var t = 0; t < o.length - 2; t += 3) {
var e = o.charAt(t + 2);
e = e >= "a" ? e.charCodeAt(0) - 87 : Number(e),
e = "+" === o.charAt(t + 1) ? r >>> e : r << e,
r = "+" === o.charAt(t) ? r + e & 4294967295 : r ^ e
}
return r
}
Copy var i = "320305.131321201" 我们运行代码，yes 成功，这样我们就完成了一次简单的 js 逆向尝试。
