试图将换行符嵌入bash中的变量中

Question

我有

var="a b c"
for i in $var
do
   p=`echo -e $p'\n'$i`
done
echo $p

我要打印最后一个回声

a
b
c

注意，我希望变量p包含换行符。我怎么做？

Answer 1

1. 插入 \n

   p="${var1}\n${var2}"
   

   echo -e “${p}”

2. 在源代码中插入新行

   p="${var1}
   

   ${var2}”
   echo “${p}”
   

3. 使用$'\n'（仅bash和zsh）

   p="${var1}"$'\n'"${var2}"
   

   echo “${p}”

细节

1.插入 \n

p="${var1}\n${var2}"
echo -e "${p}"

echo -e将两个字符解释"\n"为换行符。

var="a b c"
first_loop=true
for i in $var
do
   p="$p\n$i"            # Append
   unset first_loop
done
echo -e "$p"             # Use -e

避免多余的换行符

var="a b c"
first_loop=1
for i in $var
do
   (( $first_loop )) &&  # "((...))" is bash specific
   p="$i"            ||  # First -> Set
   p="$p\n$i"            # After -> Append
   unset first_loop
done
echo -e "$p"             # Use -e

使用功能

embed_newline()
{
   local p="$1"
   shift
   for i in "$@"
   do
      p="$p\n$i"         # Append
   done
   echo -e "$p"          # Use -e
}

var="a b c"
p=$( embed_newline $var )  # Do not use double quotes "$var"
echo "$p"

2.在源代码中插入新行

var="a b c"
for i in $var
do
   p="$p
$i"       # New line directly in the source code
done
echo "$p" # Double quotes required
          # But -e not required

避免多余的换行符

var="a b c"
first_loop=1
for i in $var
do
   (( $first_loop )) &&  # "((...))" is bash specific
   p="$i"            ||  # First -> Set
   p="$p
$i"                      # After -> Append
   unset first_loop
done
echo "$p"                # No need -e

使用功能

embed_newline()
{
   local p="$1"
   shift
   for i in "$@"
   do
      p="$p
$i"                      # Append
   done
   echo "$p"             # No need -e
}

var="a b c"
p=$( embed_newline $var )  # Do not use double quotes "$var"
echo "$p"

3.使用$'\n'（便携式性较低）

bash和zsh解释$'\n'为换行符。

var="a b c"
for i in $var
do
   p="$p"$'\n'"$i"
done
echo "$p" # Double quotes required
          # But -e not required

避免多余的换行符

var="a b c"
first_loop=1
for i in $var
do
   (( $first_loop )) &&  # "((...))" is bash specific
   p="$i"            ||  # First -> Set
   p="$p"$'\n'"$i"       # After -> Append
   unset first_loop
done
echo "$p"                # No need -e

使用功能

embed_newline()
{
   local p="$1"
   shift
   for i in "$@"
   do
      p="$p"$'\n'"$i"    # Append
   done
   echo "$p"             # No need -e
}

var="a b c"
p=$( embed_newline $var )  # Do not use double quotes "$var"
echo "$p"

所有输出都相同

a
b
c

特别感谢以下答案的贡献者：kevinf，GordonDavisson，l0b0，Dolda2000和Tripleee。

编辑

• 另请参见BinaryZebra的答案，其中提供了许多详细信息。
• Abhijeet Rastogi的答案和Dimitry的答案说明了如何避免for上述bash片段中的循环。