Mifare Classic 1K的锁定机制 [英] Locking mechanism of Mifare Classic 1K

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本文介绍了Mifare Classic 1K的锁定机制的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

Mifare Classic 1K的程序是

The procedure of Mifare Classic 1K is


  1. 轮询标签

  2. 验证这些标签

  3. 如果验证成功则读/写。

我已经完成了这些程序并且还阅读了从特定部门写入数据。

I already completed those procedures and also read and write data from specific sectors.

轮询标签的命令是

new byte[] { (byte) 0xFF, (byte) 0x00, (byte) 0x00,
                (byte) 0x00, (byte) 0x04, (byte) 0xD4, (byte) 0x4A,
                (byte) 0x01, (byte) 0x00 }

验证命令

new byte[] { (byte) 0xFF, (byte) 0x86, (byte) 0x00,
                (byte) 0x00, (byte) 0x05, (byte) 0x01,(byte) 0x00, (byte) 0x04, 
                                    (byte) 0x60,(byte) 0x00 };

这里(字节)0x01是Sector 1

Here "(byte) 0x01" is the Sector 1

写入第1区,第5区

new byte[] { (byte) 0xFF, (byte) 0x00, (byte) 0x00,(byte) 0x00, (byte) 0x15, (byte) 0xD4,
             (byte)    0x40,(byte) 0x01, (byte) 0xA0, (byte) 0x05,(byte) 0x01, (byte) 0x02, 
             (byte) 0x03,(byte) 0x04, (byte) 0x05, (byte) 0x06, (byte) 0x07,(byte) 0x08, 
             (byte) 0x09,(byte) 0x0A,(byte) 0x0B, (byte) 0x0C, (byte) 0x0D,(byte) 0x0E, 
             (byte) 0x0F, (byte) 0x10};

此处

(byte) 0x01, (byte) 0x02, (byte) 0x03,(byte) 0x04, (byte) 0x05, (byte) 0x06, (byte) 0x07,(byte) 0x08,(byte) 0x09,(byte) 0x0A,(byte) 0x0B, (byte) 0x0C, (byte) 0x0D,(byte) 0x0E,(byte) 0x0F,(byte) 0x10

是第1区第5区写的数据。

is data those are writing on block 5 on Sector 1.

从第1区和第5区读取命令是

new byte[] { (byte) 0xFF, (byte) 0x00, (byte) 0x00,
                (byte) 0x00, (byte) 0x05, (byte) 0xD4, (byte) 0x40,
                (byte) 0x01, (byte) 0x30, (byte) 0x05 };

我的相关完整代码在这里......

我的问题是如何锁定/只读特定扇区的块?

My Problem is how can I "Lock/make read only" a block from a specific sector?

推荐答案

MIFARE卡每个扇区的验证密钥和访问条件位于该扇区的最后一个块中(部门预告片)。您可以使用常规写入命令使用新访问条件和身份验证密钥更新此块。

The authentication keys and the access conditions for each sector of a MIFARE card are located in the last block of that sector (the sector trailer). You can update this block with new access conditions and authentication keys using a regular write command.

扇区预告片如下所示:

+-----------------------------+--------------+----+-----------------------------+
|  0 |  1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 | 10 | 11 | 12 | 13 | 14 | 15 |
+-----------------------------+--------------+----+-----------------------------+
|            Key A            | Access Bits  | GP |            Key B            |
|          (6 bytes)          |  (3 bytes)   | B  |          (6 bytes)          |
+-----------------------------+--------------+----+-----------------------------+

因此访问位位于字节6-8中,如下所示:

So the access bits are located in byte 6-8 and look like this:

        +-------+-------+-------+-------+-------+-------+-------+-------+
        | Bit 0 | Bit 1 | Bit 2 | Bit 3 | Bit 4 | Bit 5 | Bit 6 | Bit 7 |
        +-------+-------+-------+-------+-------+-------+-------+-------+
Byte 6: | nC2_3 | nC2_2 | nC2_1 | nC2_0 | nC1_3 | nC1_2 | nC1_1 | nC1_0 |
        +-------+-------+-------+-------+-------+-------+-------+-------+
Byte 7: |  C1_3 |  C1_2 |  C1_1 |  C1_0 | nC3_3 | nC3_2 | nC3_1 | nC3_0 |
        +-------+-------+-------+-------+-------+-------+-------+-------+
Byte 8: |  C3_3 |  C3_2 |  C3_1 |  C3_0 |  C2_3 |  C2_2 |  C2_1 |  C2_0 |
        +-------+-------+-------+-------+-------+-------+-------+-------+

其中nCx_y =不是Cx_y,C1_x,C2_x,C3_x是块x的访问条件:

Where nCx_y = not Cx_y and "C1_x, C2_x, C3_x" is the access condition for block x:


  • C1_3,C2_3,C3_3:扇区预告片(该部门的第3块)

  • C1_2,C2_2,C3_2:该部门的第2块

  • C1_1,C2_1,C3_1:该部门的第1块

  • C1_0,C2_0,C3_0:此部门中的第0块

  • C1_3, C2_3, C3_3: sector trailer (block 3 in this sector)
  • C1_2, C2_2, C3_2: block 2 in this sector
  • C1_1, C2_1, C3_1: block 1 in this sector
  • C1_0, C2_0, C3_0: block 0 in this sector

您可以找到详细信息MIFARE 1K数据表中可能的访问条件列表: http://www.nxp.com/ documents / data_sheet / MF1S503x.pdf

You can find a detailed list of possible access conditions in the MIFARE 1K datasheet: http://www.nxp.com/documents/data_sheet/MF1S503x.pdf

这篇关于Mifare Classic 1K的锁定机制的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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