这是捕获的数据包数据的示例
00000000 00 6e 0b 00 .n..
00000004 4d 5a e8 00 00 00 00 5b 52 45 55 89 e5 81 c3 81 MZ.....[ REU.....
00000014 12 00 00 ff d3 89 c3 57 68 04 00 00 00 50 ff d0 .......W h....P..
00000024 68 f0 b5 a2 56 68 05 00 00 00 50 ff d3 00 00 00 h...Vh.. ..P.....
00000034 00 00 00 00 00 00 00 00 00 00 00 00 e0 00 00 00 ........ ........
00000044 0e 1f ba 0e 00 b4 09 cd 21 b8 01 4c cd 21 54 68 ........ !..L.!Th
00000054 69 73 20 70 72 6f 67 72 61 6d 20 63 61 6e 6e 6f is progr am canno
00000064 74 20 62 65 20 72 75 6e 20 69 6e 20 44 4f 53 20 t be run in DOS
00000074 6d 6f 64 65 2e 0d 0d 0a 24 00 00 00 00 00 00 00 mode.... $.......
我只想要像这样的十六进制部分
00 6e 0b 00
4d 5a e8 00 00 00 00 5b 52 45 55 89 e5 81 c3 81
12 00 00 ff d3 89 c3 57 68 04 00 00 00 50 ff d0
我尝试右击数据包并选择复制->字节->hex流,但我得到的十六进制数据看起来根本不像上面的数据,所以我如何从wireshark复制捕获的数据包的十六进制数据?
感谢您的阅读
发布于 2015-04-01 16:39:45
在Wireshark "packet list“面板上,右键单击所需的数据包,然后:
1)如果您选择Copy->Bytes->Hex stream,您将获得一个没有空格的长字符串形式的十六进制数字
39cb08004528053f000000006f1105faac11745dac11740c039......
2)如果选择Copy->Bytes->Offset Hex,则会得到图形用户界面上显示的十六进制数字,包括每行起始字节的偏移量(帧偏移量)
0010 05 3f 00 00 00 00 6f 11 05 fa ac 11 74 5d ac 11
0020 74 0c 03 9e 03 9d 05 2b 00 00 07 e0 8f ee 8f 1c
0030 ff 00 00 00 00 00 09 0f 00 58 39 cb 60 00 00 00
0040 11 80 08 00 73 00 02 44 00 00 00 00 03 dd de de
https://stackoverflow.com/questions/29116644
复制相似问题