如何只打印hexdump中没有行号或ASCII表的十六进制值？ [重复]

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How to create a hex dump of file containing only the hex characters without spaces in bash? 5 answers
如何创建一个只包含bash中没有空格的十六进制字符的文件的十六进制转储？ 5个答案

following Convert decimal to hexadecimal in UNIX shell script

在UNIX shell脚本中将以下内容转换为十六进制

I am trying to print only the hex values from hexdump, i.e. don't print the lines numbers and the ASCII table.

我试图只打印hexdump中的十六进制值，即不打印行号和ASCII表。

But the following command line doesn't print anything:

但是以下命令行不会打印任何内容：

hexdump -n 50 -Cs 10 file.bin |  awk '{for(i=NF-17; i>2; --i) print $i}'

4 个解决方案

#1

You can specify the exact format that you want hexdump to use for output, but it's a bit tricky. Here's the default output, minus the file offsets:

您可以指定希望hexdump用于输出的确切格式，但这有点棘手。这是默认输出，减去文件偏移量：

hexdump -e '16/1 "%02x " "\n"' file.bin

(To me, it looks like this would produce an extra trailing space at the end of each line, but for some reason it doesn't.)

（对我来说，看起来这会在每一行的末尾产生额外的尾随空格，但由于某种原因，它不会。）

#2

Using xxd is better for this job:

使用xxd对于这项工作更好：

xxd -p -l 50 -seek 10 file.bin

From man xxd:

来自man xxd：

-p | -ps | -postscript | -plain
    output in postscript continuous hexdump style. Also known as plain hexdump style.

-l len | -len len
    stop after writing <len> octets.

#3

As an alternative, consider using xxd -p file.bin.

作为替代方案，请考虑使用xxd -p file.bin。

#4

First of all, remove -C which is emitting the ascii information.

首先，删除发出ascii信息的-C。

Then you could drop the offset with

然后你可以删除偏移量

hexdump -n 50 -s 10 file.bin | cut -c 9-

#1