如何为bash变量分配多行值 [英] How to assign a multiple line value to a bash variable

问题描述

我有一个变量 FOO ，需要为它分配一个多行值.像这样的东西

I have a variable FOO with me that needs to be assigned with a value that will be multiple lines. Something like this,

FOO="This is line 1
     This is line 2
     This is line 3"

因此，当我打印 FOO 的值时，应提供以下输出.

So when I print the value of FOO it should give the following output.

echo $FOO
output:
This is line 1
This is line 2
This is line 3

此外，行数将动态确定，因为我将使用循环对其进行初始化.

Furthermore, the number of lines will be decided dynamically as I will initialize it using a loop.

在另一个问题中主要使用 read -d 显示的答案不适合我，因为我正在执行大量的字符串操作，并且代码格式也很重要.

The answers that have been shown in the other question using mainly read -d is not suitable for me as I am doing intensive string operations and the code format is also important.

推荐答案

不要缩进行，否则您会得到多余的空格.展开"$ FOO" 时，请使用引号，以确保保留换行符.

Don't indent the lines or you'll get extra spaces. Use quotes when you expand "$FOO" to ensure the newlines are preserved.

$ FOO="This is line 1 
This is line 2   
This is line 3"
$ echo "$FOO"
This is line 1
This is line 2
This is line 3

另一种方法是使用 \ n 转义序列.它们在 $'...'字符串内解释.

Another way is to use \n escape sequences. They're interpreted inside of $'...' strings.

$ FOO=$'This is line 1\nThis is line 2\nThis is line 3'
$ echo "$FOO"

第三种方法是存储字符 \ 和 n ，然后让 echo -e 解释转义序列.这是一个微妙的区别.重要的是， \ n 不会在常规引号内进行解释.

A third way is to store the characters \ and n, and then have echo -e interpret the escape sequences. It's a subtle difference. The important part is that \n isn't interpreted inside of regular quotes.

$ FOO='This is line 1\nThis is line 2\nThis is line 3'
$ echo -e "$FOO"
This is line 1
This is line 2
This is line 3

如果删除 -e 选项并让 echo 打印原始字符串而不解释任何内容，则可以看到我的区别.

You can see the distinction I'm making if you remove the -e option and have echo print the raw string without interpreting anything.

$ echo "$FOO"
This is line 1\nThis is line 2\nThis is line 3

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