用换行符在 bash 中构建一个字符串
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Build a string in bash with newlines
提问by Palace Chan
I'm going through a script in bash where depending on the conditionals I want to append to a variable different things and then display it at the very end, something like this:
我正在 bash 中浏览一个脚本，其中根据我想附加到变量的不同内容的条件，然后在最后显示它，如下所示：
VAR="The "
if [[ whatever ]]; then
VAR="$VAR cat wears a mask"
elif [[ whatevs ]]; then
VAR="$VAR rat has a flask"
fi
but i run into difficulties if I try to use this form of building up VAR by appending to it when I want to occasionally append newlines into it. How would I do VAR="$VAR\nin a box", for example? I have seen usage of $'\n'before but not when also trying to use $VARbecause of the appending.
但是，如果我想偶尔在其中添加换行符时，尝试通过附加到它来使用这种构建 VAR 的形式，我会遇到困难。VAR="$VAR\nin a box"例如，我该怎么做？我之前看到过使用，$'\n'但$VAR由于附加而在尝试使用时没有看到。
回答by gniourf_gniourf
Using ANSI-C quoting:
使用ANSI-C 引用：
var="$var"$'\n'"in a box"
You could put the $'\n'in a variable:
你可以把它$'\n'放在一个变量中：
newline=$'\n'
var="$var${newline}in a box"
By the way, in this case, it's better to use the concatenation operator:
顺便说一下，在这种情况下，最好使用连接运算符：
var+="${newline}in a box"
If you don't like ANSI-C quoting, you can use printfwith its -voption:
如果你不喜欢 ANSI-C 引用，你可以使用printf它的-v选项：
printf -v var '%s\n%s' "$var" "in a box"
Then, to print the content of the variable var, don't forget quotes!
然后，要打印变量的内容var，不要忘记引号！
echo "$var"
or, better yet,
或者，更好的是，
printf '%s\n' "$var"
Remark.Don't use upper case variable names in Bash. It's terrible, and one day it will clash with an already existing variable!
评论。不要在 Bash 中使用大写变量名。这太可怕了，总有一天它会与一个已经存在的变量发生冲突！
You could also make a function to append a newline and a string to a variable using indirect expansion (have a look in the Shell Parameter Expansionsection of the manual) as so:
您还可以创建一个函数，使用间接扩展（查看手册的Shell 参数扩展部分）将换行符和字符串附加到变量，如下所示：
append_with_newline() { printf -v "" '%s\n%s' "${!1}" ""; }
Then:
然后：
$ var="The "
$ var+="cat wears a mask"
$ append_with_newline var "in a box"
$ printf '%s\n' "$var"
The cat wears a mask
in a box
$ # there's no cheating, look at the content of var:
$ declare -p var
declare -- var="The cat wears a mask
in a box"
Just for fun, here's a generalized version of the append_with_newlinefunction that takes n+1arguments (n≥1) and that will concatenate them all (with exception of the first one being the name of a variable that will be expanded) using a newline as separator, and puts the answer in the variable, the name of which is given in the first argument:
只是为了好玩，这里有一个通用版本的append_with_newline函数，它接受n+1 个参数（n≥1）并使用换行符将它们全部连接起来（第一个是将被扩展的变量的名称除外）作为分隔符，并将答案放在变量中，变量的名称在第一个参数中给出：
concatenate_with_newlines() { local IFS=$'\n'; printf -v "" '%s\n%s' "${!1}" "${*:2}"; }
Look how well it works:
看看效果如何：
$ var="hello"
$ concatenate_with_newlines var "a gorilla" "a banana" "and foobar"
$ printf '%s\n' "$var"
hello
a gorilla
a banana
and foobar
$ # :)
It's a funny trickery with IFSand "$*".
这是一个有趣的权谋与IFS和"$*"。
回答by chepner
By default, bashdoes not process escape characters. The assignment
默认情况下，bash不处理转义字符。那作业
VAR="foo\nbar"
assigns 8 characters to the variable VAR: 'f', 'o', 'o', '\', 'n', 'b', 'a', and 'r'. The POSIX standard states that the echocommand should treat the two-character string \nin its arguments as a linefeed; bashdoes not follow the standard in this case and requires the -eoption to enable this processing. The printfcommand follows the POSIX specification and treats literal "\n" in its argument as a linefeed, but does not expand such uses in strings that replace placeholders: printf "%s\n" "$VAR"would still output
为变量分配 8 个字符VAR：'f'、'o'、'o'、'\'、'n'、'b'、'a' 和 'r'。POSIX 标准规定该echo命令应将\n其参数中的两个字符的字符串视为换行符；bash在这种情况下不遵循标准，并且需要-e启用此处理的选项。该printf命令遵循 POSIX 规范并将其参数中的文字“\n”视为换行符，但不会在替换占位符的字符串中扩展此类用途：printf "%s\n" "$VAR"仍会输出
foo\nbar
instead of
代替
foo
bar
To include an actual linefeed character in a string, you can use ANSI quoting:
要在字符串中包含实际的换行符，您可以使用 ANSI 引用：
VAR=$'foo\nbar'
which has the drawback that the string is otherwise processed as a single-quoted string, and cannot contain parameter expansions or command substitutions. Another option is that a string that spans multiple lines will contain the quoted linefeed characters:
它的缺点是字符串被处理为单引号字符串，并且不能包含参数扩展或命令替换。另一种选择是跨多行的字符串将包含带引号的换行符：
$ VAR="foo
> bar"
$ echo "$foo"
foo
bar
回答by anubhava
It works this way:
它是这样工作的：
> VAR=foo
> VAR="$VAR\nrat has a flask"
> echo -e "$VAR"
foo
rat has a flask
