bash 从bash中的文件读取的行中删除换行符
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removing linebreak from a line read from a file in bash
提问by user3304726
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($Line) ) in shell script.
我有一个包含一些行的文件。我想将每一行存储到一个变量中，但该行必须在 shell 脚本中被 chomped（在 perl 中完成的方式 - chomp($Line) ）。
The code containing the functionality of opening a file and reading the lines is
包含打开文件和读取行的功能的代码是
p_file() {
File=
count=0
while read LINE
do
chomped_line=$LINE **#How to delete the linebreaks**
echo $chomped_line
done < $File
}
How to delete the linebreaks and store the string in the variable(chomped_line) as above
如何删除换行符并将字符串存储在变量（chomped_line）中，如上
回答by konsolebox
Simply use
只需使用
while IFS=$' \t\n\r' read -r LINE
It would exclude leading and trailing spaces even carriage returns characters (\r) every line. No need to chomp it.
它会排除前导和尾随空格，甚至\r每行回车符 ( )。没必要咬它。
If you still want to include other spaces besides \nand/or \r, just don't specify the others:
如果您还想包含除\nand/or之外的其他空格\r，请不要指定其他空格：
while IFS=$'\n\r' read -r LINE
Another way if you don't like using IFS is just to trim out \r:
如果您不喜欢使用 IFS，另一种方法就是修剪\r：
chomped_line=${LINE%$'\r'}
* -rprevents backslashes to escape any characters.
* -r防止反斜杠转义任何字符。
