如何在unix中删除文件的最后一个字符? [英] How can I remove the last character of a file in unix?

问题描述

假设我有一些任意的多行文本文件:

Say I have some arbitrary multi-line text file:

sometext
moretext
lastline

如何在不使文本文件无效的情况下仅删除文件的最后一个字符(e，而不是换行符或空字符)?

How can I remove only the last character (the e, not the newline or null) of the file without making the text file invalid?

推荐答案

更简单的方法(输出到标准输出，不更新输入文件):

A simpler approach (outputs to stdout, doesn't update the input file):

sed '$ s/.$//' somefile

• $ 是只匹配最后输入行的 Sed 地址，从而导致执行以下函数调用(s/.$//)仅在最后一行.

  • $ is a Sed address that matches the last input line only, thus causing the following function call (s/.$//) to be executed on the last line only.

    s/.$// 用空字符串替换(在本例中为 last)行的最后一个字符；即，有效地删除最后一个字符.(换行前)就行了.
    ^{. 匹配该行中的任何字符，紧随其后的 $ 将匹配锚定到该行的末尾；请注意 $ 在这个 正则表达式 中的使用在概念上是如何相关的，但在技术上与之前将 $ 用作 Sed 不同地址.}

    s/.$// replaces the last character on the (in this case last) line with an empty string; i.e., effectively removes the last char. (before the newline) on the line.
    ^{. matches any character on the line, and following it with $ anchors the match to the end of the line; note how the use of $ in this regular expression is conceptually related, but technically distinct from the previous use of $ as a Sed address.}

    stdin 输入示例(假设使用 Bash、Ksh 或 Zsh):

    Example with stdin input (assumes Bash, Ksh, or Zsh):

      $ sed '$ s/.$//' <<< $'line one
    line two'
      line one
      line tw
    

  也更新输入文件(如果输入文件是符号链接，则不要使用):

  To update the input file too (do not use if the input file is a symlink):

  sed -i '$ s/.$//' somefile
  

  ^注意:

  • 在 macOS 上，您必须使用 -i '' 而不是 -i；有关与 -i 相关的陷阱的概述，请参阅此答案的下半部分.
  • 如果您需要处理非常大的输入文件和/或性能/磁盘使用是一个问题并且您正在使用 GNU 实用程序 (Linux)，请参阅ImHere 的有用答案.
  • On macOS, you'd have to use -i '' instead of just -i; for an overview of the pitfalls associated with -i, see the bottom half of this answer.
  • If you need to process very large input files and/or performance / disk usage are a concern and you're using GNU utilities (Linux), see ImHere's helpful answer.

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