linux – 当分割成多行时,如何避免echo中的空格

我有一个很长的字符串由echo命令打印.通过这样做,我希望它完美缩进.
我正在尝试这个,它完美地工作

echo "This is a very long string. An"
"d it is printed in one line"

Output:
This is a very long string. And it is printed in one line

但是当我尝试缩进它时,echo语句也缩进了.它增加了额外的空间.

echo "This is a very long string. An"
    "d it is printed in one line"

Output:
This is a very long string. An d it is printed in one line

我找不到任何可以完美做到这一点的工作反应.

解决方法

$echo "a"             "b"
a b
$echo "a" "b"
a b
$echo "a"
>           "b"
a b

如果要完全控制要打印的内容,请使用带有printf的数组：

lines=("This is a very long string. An"
       "d it is printed in one line")
printf "%s" "${lines[@]}"
printf "n"

这将返回：

This is a very long string. And it is printed in one line

或者作为suggested by 123 in comments,使用echo也将数组设置IFS设置为null：

# we define the same array $lines as above

$IFS=""
$echo "${lines[*]}"
This is a very long string. And it is printed in one line
$unset IFS
$echo "${lines[*]}"
This is a very long string. An d it is printed in one line
#                             ^
#                             note the space

从Bash manual → 3.4.2. Special Parameters开始：

*

($) Expands to the positional parameters,starting from one. When the expansion is not within double quotes,each positional parameter expands to a separate word. In contexts where it is performed,those words are subject to further word splitting and pathname expansion. When the expansion occurs within double quotes,it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is,“$” is equivalent to “$1c$2c…”,where c is the first character of the value of the IFS variable. If IFS is unset,the parameters are separated by spaces. If IFS is null,the parameters are joined without intervening separators.

有趣的阅读：Why is printf better than echo?.

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