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  • python实现Bencode解码方法

         近期搞项目中遇到Bencode解码的问题,就用Py写了个Bencode解码的代码。作为笔记保存参考。

      BEncoding是BitTorrent用在传输数据结构的编码方式,这种编码方式支持四种类型的数据:string, int, Dictionary<string, object>, List<object>,各自的编码规则如下:

      • string类型的编码格式为[length]:[string]。以字符串的长度开头,加一个冒号,并以字符串内容结束。示例:"abc" => 3:abc
      • int类型的编码格式为i[int]e。以i开头,加上数字,以e结尾。 示例:123 => i123e
      • List<object>类型的编码格式为l[object]e。以l开头,加上列表中各个元素的编码(元素的类型同样为BEncoding支持的类型),以e结尾。 示例:List<"abc", 123> => l3:abci123ee
      • Dictionary<string, object>类型的编码格式为d[Key-Value Pair]e。以d开头,加上字典中每个键值对的编码,以e结尾。
    # 解析List<object>类型,例:l3:abci123ee => List<"abc", 123>
    def decode_list(str,p):
        l = len(str)
        # print(l)
        # print(p)
        list = []
        while p < l:
            if str[p + 1].isdigit():
                f_m = str.index(":",p + 1,l)
                f_n = int(str[p + 1:f_m])
                f_e = f_m + f_n
                data_str = str[f_m + 1:f_e + 1]
                list.append(data_str)
                p = f_e
            elif str[p + 1] == "i":
                end = str.index("e", p + 1, l)
                data_int = str[p + 2: int(end)]
                list.append(data_int)
                p = end
            elif str[p + 1] == "e":
                p = p + 2
                list.append("p")
                list.append(p)
                break
            elif str[p + 1] == "l":
                p = p + 1
        return list
     
    # 解析Dictionary<string, object>类型,例:d4:name11:create chen3:agei23ee => Dictionary<{"name":"create chen"},{"age":23}>
    def decode_dict(str,p):
        test = decode_list(str,p)
     
        data_seq = []
        data_val = []
     
        i = 0
        while i < len(test):
            data_seq.append(test[i])
            data_val.append(test[i + 1])
            i = i + 2
     
        data_dict = dict.fromkeys(data_seq)
     
        i = 0
        while i < len(test) / 2:
            data_dict[data_seq[i]] = data_val[i]
            i = i + 1
     
        return data_dict
     
    data = "d4:name11:create chen3:agei23eel3:abci123eei23e3:age"
    print(data)
    # 定义全局指针,追踪所查位置
    global p
    l = len(data)
    p = 0
    while p < l:
        # 判断是否为dict型
        if data[p] == "d":
            data_dict = decode_dict(data,p)
            p = int(data_dict["p"])
            del data_dict["p"]
            print(data_dict)
            f = open("text.txt","a+")
            for k, v in data_dict.items():
                f.write(k + ":" + str(v))
                f.write("
    ")
            f.close()
        # 判断是否为list型
        elif data[p] == "l":
            data_list = decode_list(data,p)
            p = int(data_list[-1])
            del data_list[-1]
            del data_list[-1]
            print(data_list)
            f = open("text.txt", "a+")
            for i in data_list:
                f.write(i)
                f.write("
    ")
            f.close()
        # 判断是否为str型,解析
        elif data[p] == "i":
            f = data.index("e",p,l)
            data_int = data[p + 1:f]
            p = f + 1
            print(data_int)
            f = open("text.txt", "a+")
            f.write(data_int)
            f.write("
    ")
            f.close()
        # 判断是否为int型,解析(为了方便,暂未将其结果转为int型,而是str型,例:“123”
        elif data[p].isdigit():
            f = data.index(":",p,l)
            print(data[f])
            data_str = data[f + 1:int(data[p:f]) + f + 1]
            print(data_str)
            p = int(data[p:f]) + f + 1
            f = open("text.txt", "a+")
            f.write(data_str)
            f.write("
    ")
            f.close()
    print("It's done")
    

      

      

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  • 原文地址:https://www.cnblogs.com/68xi/p/9294309.html
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