如何在sh中的字符串中添加换行符？

Question

这个

STR="Hello\nWorld"
echo $STR

产生输出

Hello\nWorld

代替

Hello
World

如何在字符串中添加换行符？

注意:这个问题不是关于回声. 我知道echo -e,但我正在寻找一种解决方案,允许将一个字符串(包括换行符)作为参数传递给其他命令,这些命令没有类似的选项来解释\n为换行符.

Answer 1

解决方案是使用$'string',例如:

$ STR=$'Hello\nWorld'
$ echo "$STR"
Hello
World

以下是Bash手册页的摘录:

   Words of the form $'string' are treated specially.  The word expands to
   string, with backslash-escaped characters replaced as specified by  the
   ANSI  C  standard.  Backslash escape sequences, if present, are decoded
   as follows:
          \a     alert (bell)
          \b     backspace
          \e
          \E     an escape character
          \f     form feed
          \n     new line
          \r     carriage return
          \t     horizontal tab
          \v     vertical tab
          \\     backslash
          \'     single quote
          \"     double quote
          \nnn   the eight-bit character whose value is  the  octal  value
                 nnn (one to three digits)
          \xHH   the  eight-bit  character  whose value is the hexadecimal
                 value HH (one or two hex digits)
          \cx    a control-x character

   The expanded result is single-quoted, as if the  dollar  sign  had  not
   been present.

   A double-quoted string preceded by a dollar sign ($"string") will cause
   the string to be translated according to the current  locale.   If  the
   current  locale  is  C  or  POSIX,  the dollar sign is ignored.  If the
   string is translated and replaced, the replacement is double-quoted.

Answer 2

Echo是九十年代,充满了危险,它的使用应该导致核心转储不低于4GB.说真的,echo的问题是Unix标准化过程最终发明printf实用程序的原因,消除了所有问题.

所以要在字符串中获取换行符:

FOO="hello
world"
BAR=$(printf "hello\nworld\n") # Alternative; note: final newline is deleted
printf '<%s>\n' "$FOO"
printf '<%s>\n' "$BAR"

那里!没有SYSV与BSD的回声疯狂,一切都得到了整齐的打印和完全便携式支持C转义序列.大家请printf现在使用,永远不要回头.

Answer 3

我根据其他答案做了什么

NEWLINE=$'\n'
my_var="__between eggs and bacon__"
echo "spam${NEWLINE}eggs${my_var}bacon${NEWLINE}knight"

# which outputs:
spam
eggs__between eggs and bacon__bacon
knight

Answer 4

问题不在于shell.问题实际上是echo命令本身,以及变量插值周围缺少双引号.您可以尝试使用echo -e但并非在所有平台上都支持,printf现在建议使用其中一个原因来实现可移植性.

您也可以尝试将换行符直接插入到shell脚本中(如果脚本是您正在编写的脚本),那么它看起来像......

#!/bin/sh
echo "Hello
World"
#EOF

或者等价的

#!/bin/sh
string="Hello
World"
echo "$string"  # note double quotes!

Answer 5

1. 唯一简单的替代方法是在变量中实际输入一个新行:

   $ STR='new
   line'
   $ printf '%s' "$STR"
   new
   line
   

   是的,这意味着Enter在代码中需要编写代码.

2. new line角色有几个等价物.

   \n           ### A common way to represent a new line character.
   \012         ### Octal value of a new line character.
   \x0A         ### Hexadecimal value of a new line character.
   

   但所有这些都需要一些工具(POSIX printf)的"解释" :

   echo -e "new\nline"           ### on POSIX echo, `-e` is not required.
   printf 'new\nline'            ### Understood by POSIX printf.
   printf 'new\012line'          ### Valid in POSIX printf.
   printf 'new\x0Aline'       
   printf '%b' 'new\0012line'    ### Valid in POSIX printf.
   

   因此,该工具需要使用换行符构建一个字符串:

   $ STR="$(printf 'new\nline')"
   $ printf '%s' "$STR"
   new
   line
   

3. 在某些shell中,序列$'是一个特殊的shell扩展.已知可以在ksh93,bash和zsh中工作:

   $ STR=$'new\nline'
   

4. 当然,更复杂的解决方案也是可能的:

   $ echo '6e65770a6c696e650a' | xxd -p -r
   new
   line
   

   要么

   $ echo "new line" | sed 's/ \+/\n/g'
   new
   line
   

Answer 6

我发现-e国旗优雅而直截了当

-e#outputs-e

如果字符串是另一个命令的输出,我只使用引号

-e

Answer 7

单引号前面的$如下所示，但是双引号无效。

$ echo $'Hello\nWorld'
Hello
World
$ echo $"Hello\nWorld"
Hello\nWorld