从文本文件中回显第 n 行，其中“n"是命令行参数

问题描述

我有一个带有数字的简单文本文件，例如:

I have a simple text file with numbers like:

12345
45678
34567
89101

我需要一个批处理来返回这个文件的第 n 行.n 应该取自命令行参数.

I need a batch that will return the nth line from this file. n should be taken from a command line argument.

我对批处理脚本很陌生，所以在此先感谢您的帮助.

I am very new to batch scripting so Thanks in advance for any help on this.

推荐答案

要获取文件来自的第 n 行，您可以使用 more +n (For line1 is n=0).
要拆分文件的其余部分，您可以使用 FOR/F 循环.

To get the file from the nth line you could use more +n (For line1 is n=0).
To split the rest of the file you could use a FOR /F loop.

即使在第 n 行之前有空行，这也有效.
可能需要将 EOL 设置为未使用的字符或换行(默认为 ;)

This works even, if there are empty lines before the nth line.
It could be necessary to set the EOL to an unused character or to linefeed (default is ;)

set "lineNr=%1"
set /a lineNr-=1
for /f "usebackq delims=" %%a in (`more +%lineNr% text.txt`) DO (
  echo %%a
  goto :leave
)
:leave

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