网络爬虫-网易易盾滑块验证码
首先打开易盾官网，依次单击“在线体验”和“滑块拼图”选项，出现如下图所示的滑块验证码，滑动后发现Network面板成功抓包。
如图所示 其中主要的加密参数可以看出来就是acToken以及cb 我们接着下断点往下走
我通过上图可以看到 acToken，同时也发现了data，这里 data 先放一下，我们追溯 acToken ，这里我们就看栈，通过栈追溯到它。
这里我是在箭头指的这个栈对应的位置打的断点。这里我们再次让网页运行，当断点断到这里时，可以看到当前栈之上的栈没有 acToken 类似样式的值出现，那就说明，acToken 是在这里或者之后出现的，我们先分析当前的，看看是否是，如若不是再继续向下找。这里我们进入上面函数 cc
这里我们在 开头 以及两个 return 处打上断点，这里我们让它自己运行到return处，从下往上分析。
可以看到这里生成的值跟上面看到的 acToken 相似，同时这里也说明了 acToken 在这里生成的。
还有一个要注意的是函数cc里面的参数b是，上图api请求返回回来的dt值。
剩下就扣代码吧，细心点就可以搞出来。
下面我们逆向参数 data，其实参数 data 本质上就是滑块轨迹的加密。
这里我们可以在上面一开始找 acToken 时发现有 data，我们在那里打上断点，然后再追栈，最后会发现在下图发现 data，再往上追，就找不到了
那我们就找一下 e ,但是 e 太多了，不好找，那我们找 data ，但是 data 也有点断，也不到找，这里我们要根据 data 的特征下手了。
我把一个data复制下来，然后序列化一下，这里我是通过 ext 找的。
最后找到一个，这里可以看到跟上面序列化得到的差不多，那我们打一下断点看看。
可以看到已经成功断到这里了，那也说明了 data 在这里生成，
上面 this.traceData 就是加密的轨迹，u 其实是个定值 50 。
这里我们搜索一下 this.traceData 。可以看到跟滑块的x和y的坐标还有时间有关。最终用f函数做了一下加密处理。我们这里不方便直接断点。因为，在此处断点，每动一下鼠标都要debug。这里我们可以通过条件断点打印一下。
Math.round(e.clientX - i) 为第一个是横坐标的拖动距离，
Math.round(e.clientY - n) 为第二个是纵坐标上下浮动的距离，
s.now() - this.beginTime 为第三个是当前滑动的时间-开始滑动的时间。
然后，剩下的就是一个扣代码了以及缺口识别了，缺口识别的话可以用ddddocr或者目前已经有很多的开源ocr来搞，反正能精准识别到距离就行。
这里给一个ddddocr的例子
import ddddocr
def get_gap(target_img:bytes, background_img:bytes):
det = ddddocr.DdddOcr(det=False, ocr=False, show_ad=False)
res = det.slide_match(target_img, background_img,simple_target=True)
print(res)
return res["target"][0]
下图是我本地生成的 data 值。
cb生成部分代码
function uuid(e, t) {
var i = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".split("")
, a = []
, n = void 0;
if (t = t || i.length,
e)
for (n = 0; n < e; n++)
a[n] = i[0 | Math.random() * t];
else {
var r = void 0;
for (a[8] = a[13] = a[18] = a[23] = "-",
a[14] = "4",
n = 0; n < 36; n++)
a[n] || (r = 0 | 16 * Math.random(),
a[n] = i[19 === n ? 3 & r | 8 : r])
}
return a.join("")
}
__toByte = function (e) {
function t(t) {
return e.apply(this, arguments)
}
return t.toString = function () {
return e.toString()
}
,
t
}(function (e) {
if (e < -128)
return __toByte(128 - (-128 - e));
if (e >= -128 && e <= 127)
return e;
if (e > 127)
return __toByte(-129 + e - 127);
throw new Error("1001")
});
n = function (e, t) {
return __toByte(e + t)
}
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return null;
if (null == t)
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i[o] = n(e[o], t[o % r]);
return i
}
oo = function (e, t) {
return e = __toByte(e),
t = __toByte(t),
__toByte(e ^ t)
}
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return i
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s = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f"]
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t.push(s[15 & e]),
t.join("")
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var t = e.length;
if (null == e || t < 0)
return new String("");
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i.push(l(e[n]));
return i.join("")
}
c = function (e) {
if (null == e || void 0 == e)
return e;
for (var t = encodeURIComponent(e), i = [], n = t.length, r = 0; r < n; r++)
if ("%" == t.charAt(r)) {
if (!(r + 2 < n))
throw new Error("1009");
i.push(f(t.charAt(++r) + "" + t.charAt(++r))[0])
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i.push(t.charCodeAt(r));
return i
}
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var t = [];
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t[3] = 255 & e,
t
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var t = j(e);
return u(t)
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throw new Error("1003");
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n[r] = e[t + r];
return n
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if (null == i)
throw new Error("1004");
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throw new Error("1003");
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return i
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l.push(s);
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r = e[t + 1],
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l.push(a[n >>> 2 & 63]),
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l.push(s);
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throw new Error("1010");
n = e[t],
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l.push(a[n >>> 2 & 63]),
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break
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i.push(m(e, n, t)),
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// try {
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p(e, 0, n, 0, t);
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var o = j(t);
return p(o, 0, n, t + i, O),
n
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x = function (e) {
if (null == e || e.length % C != 0)
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var i = 256 * Math.random();
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e[t] = __toByte(i)
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D = function (e, t) {
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r.push(n(e[a], i));
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function cb() {
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最终成品代码图
