在BSD中如何使用新行替换?

Greetings, how do I perform the following in BSD sed?

您好，我如何在BSD中执行以下操作?

sed 's/ /\n/g'

From the man-page it states that \n will be treated literally within a replacement string, how do I avoid this behavior? Is there an alternate?

在手册页中，它声明将在替换字符串中处理\n，我如何避免这种行为?有另一个吗?

I'm using Mac OS Snow Leopard, I may install fink to get GNU sed.

我正在使用Mac OS雪豹，我可能会安装fink以获得GNU sed。

4 个解决方案

#1

In a shell, you can do:

在壳层中，你可以做到:

    sed 's/ /\
/g'

hitting the enter key after the backslash to insert a newline.

在反斜杠后按enter键插入换行。

#2

Another way:

另一种方法:

sed -e 's/ /\'$'\n/g'

See here.

在这里看到的。

#3

For ease of use, i personally often use

为了便于使用，我个人经常使用

cr="\n" 
# or (depending version and OS)
cr="
"

sed "s/ /\\${cr}/g"

so it stays on 1 line.

它在一条直线上。

#4

To expand on @sikmir's answer: In Bash, which is the default shell on Mac OS X, all you need to do is place a $ character in front of the quoted string containing the escape sequence that you want to get interpreted. Bash will automatically translate it for you.

要扩展@sikmir的答案:在Bash中，这是Mac OS X上的默认shell，您所需要做的就是在引用的字符串前面加上一个$字符，其中包含您想要解释的转义序列。Bash将自动为您翻译它。

For example, I removed all MS-DOS carriage returns from all the source files in lib/ and include/ by writing:

例如，我删除了lib/中所有源文件的所有MS-DOS回车，并包括/通过写入:

grep -lr $'\r' lib include | xargs sed -i -e $'s/\r//'
find . -name '*-e' -delete

BSD grep would have interpreted '\r' correctly on its own, but using $'\r' doesn't hurt.

BSD grep本身就可以正确地解释“\r”，但是使用$“\r”并不会造成伤害。

BSD sed would have misinterpreted 's/\r//' on its own, but by using $'s/\r//', I avoided that trap.

BSD sed本来可以自行错误地解释's/\r/ '，但是通过使用$'s/\r/ ' '，我避免了这个陷阱。

Notice that we can put $ in front of the entire string, and it will take care of all the escape sequences in the whole string.

注意，我们可以把$放在整个字符串前面，它将处理整个字符串中的所有转义序列。

$ echo $'hello\b\\world'
hell\world

#1