linux - cut 命令忽略空格

Question

问题是:显示每行文本中的和字符。

这是我的代码:

while read line
do
    a=`echo $line | cut -c 2`
    b=`echo $line | cut -c 7`

    echo -n $a
    echo $b
done

问题是当第一个字符是空格时，它不打印空格。例如，

Input:
A big elephant

Expected output:
 e

My output:
e

如何解决这个问题？

Answer 1

您需要引用变量。当 bash 扩展 $a 时，echo 不会打印空格，因为它不被视为参数。

一个命令可以有很多尾随空格，但由于 bash 默认使用空格来分隔命令和参数，任何额外的空格都将被忽略，除非明确地成为参数的一部分(使用引号或转义)。

例如:

=> echo a       # Argument without spaces works (at least in this case)
a
=> echo  a      # Two spaces between, unquoted spaces are ignored
a
=> echo " a"    # Quotes make the space part of the argument
 a
=> echo      a  # Arguments can have many spaces between them
a
=> echo         # No quotes or spaces, echo sees no arguments, doesn't print anything

=> echo " "     # Space is printed (changed to an underscore, wouldn't actually be visible)
_
=>

这也是为什么必须在文件名中转义空格，除非它们被引用:

=> touch file 1      # Makes two files: "file" and "1"
=> touch "file 1"    # Makes one file: "file 1"
=> touch file\ 1     # Makes one file: "file 1"

您的最终代码将是:

while read line
do
    a=`echo $line | cut -c 2`
    b=`echo $line | cut -c 7`

    echo -n "$a"
    echo "$b"
done