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当我没有可用的“-z”选项时，如何执行与“head -z”等效的命令？

tag-icon • tag-icon coreutils head coreos
当我没有可用的“-z”选项时，如何执行与“head -z”等效的命令？

我需要head -z一个脚本（偏离主题，但可以找到动机在这个问题中），但在我的 CoreOS 835.13.0 中我得到head: invalid option -- 'z'.

完整head --help输出：

Usage: head [OPTION]... [FILE]...
Print the first 10 lines of each FILE to standard output.
With more than one FILE, precede each with a header giving the file name.
With no FILE, or when FILE is -, read standard input.

Mandatory arguments to long options are mandatory for short options too.
  -c, --bytes=[-]K         print the first K bytes of each file;
                             with the leading '-', print all but the last
                             K bytes of each file
  -n, --lines=[-]K         print the first K lines instead of the first 10;
                             with the leading '-', print all but the last
                             K lines of each file
  -q, --quiet, --silent    never print headers giving file names
  -v, --verbose            always print headers giving file names
      --help     display this help and exit
      --version  output version information and exit

K may have a multiplier suffix:
b 512, kB 1000, K 1024, MB 1000*1000, M 1024*1024,
GB 1000*1000*1000, G 1024*1024*1024, and so on for T, P, E, Z, Y.

GNU coreutils online help: <http://www.gnu.org/software/coreutils/>
Report head translation bugs to <http://translationproject.org/team/>
For complete documentation, run: info coreutils 'head invocation'

有趣的是，最后一行告诉我要运行，info coreutils 'head invocation'但我得到了info: command not found。

答案1

交换 head 前后的 NUL 和 NL：

<file tr '\0\n' '\n\0' | head | tr '\n\0' '\0\n'

使用最新版本的 GNU sed：

sed -z 10q

使用 GNU awk：

gawk -v RS='\0' -v ORS='\0' '{print}; NR == 10 {exit}'

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