从bash中的文件读取的行中删除换行符 [英] removing linebreak from a line read from a file in bash

问题描述

我有一个包含一些行的文件.我想将每一行存储到一个变量中，但是必须在shell脚本中将这一行切掉(在perl中完成它的方式-chomp($ Line)).

I have a file containing some lines. I wanted to store each line to a variable , but the line must be chomped (the way its done in perl - chomp($Line) ) in shell script.

包含打开文件和读取行的功能的代码为

The code containing the functionality of opening a file and reading the lines is

p_file() {

File=$1
count=0

while read LINE
do
      chomped_line=$LINE    **#How to delete the linebreaks** 
      echo $chomped_line

done < $File

}

如何删除换行符并将字符串存储在上述变量(chomped_line)中

How to delete the linebreaks and store the string in the variable(chomped_line) as above

推荐答案

只需使用

while IFS=$' \t\r\n' read -r line

即使每行回车返回字符( \ r )，也将排除前导和尾随空格.无需砍.

It would exclude leading and trailing spaces even carriage returns characters (\r) every line. No need to chomp it.

如果您仍然想在 \ n 和/或 \ r 之外添加其他空格，则不要指定其他空格:

If you still want to include other spaces besides \n and/or \r, just don't specify the others:

while IFS=$'\r\n' read -r line

如果您不喜欢使用IFS，另一种方法是修剪 \ r :

Another way if you don't like using IFS is just to trim out \r:

chomped_line=${line%$'\r'}

* -r 防止反斜杠转义任何字符.

  *  -r prevents backslashes to escape any characters.

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