爬虫js逆向实战--韵达分析

内容仅供参考学习

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目标分析：

• 首先模拟一次滑块滑动：
  

上图中被框住的参数为滑块的信息以及可以滑块的纵坐标

• 此时我们需要获取的只有x， 而获取x的方法就是计算出滑块初始位置到缺口的位置, 可以利用ddddocr：

	ocr = ddddocr.DdddOcr(det=False, ocr=False)
    timestamp_1, randomStr_1, signature_1 = get_encrypt()
    params_1 = {
        "wid": "1",
        "timeStamp": timestamp_1,
        "randomStr": randomStr_1,
        "signature": signature_1,
    }
    session.get("http://ydgw.yundasys.com:31620/index.php/api/order.record/captcha_type", params=params_1)
    time.sleep(random.uniform(0.5, 1.0))
    timestamp, randomStr, signature = get_encrypt()
    params = {
        "wid": "1",
        "timeStamp": timestamp,
        "randomStr": randomStr,
        "signature": signature,
    }
    res = session.get("http://ydgw.yundasys.com:31620/index.php/api/order.record/captcha", params=params).json()['data']
    time.sleep(random.randint(1, 2))
    big_img = res['big']
    small_img = res['small']
    x = ocr.slide_match(target_bytes=base64.b64decode(small_img), background_bytes=base64.b64decode(big_img))['target'][0]
    y = res['y']

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