替换包含CRLF字符串？ [英] Replace string that contains CRLF?

问题描述

我重新格式化文件，我想执行以下步骤：

I'm reformatting a file, and I want to perform the following steps:

1. 替换双CRLF的一个临时的字符序列（ $ CRLF $ 或东西）


2. 删除所有CRLF在整个文件


3. 回去更换双CRLF的。

所以像这样输入：

This is a paragraph
of text that has
been manually fitted
into a certain colum
width.

This is another
paragraph of text
that is the same.

将成为

This is a paragraph of text that has been manually fitted into a certain colum width.

This is another paragraph of text that is the same.

看来这应该通过一些简单的 SED 程序管道输入的可能，但我不知道如何引用 CRLF 在 SED （在使用 SED的/＆LT; CRLF＆GT;＆LT; CRLF＆GT; / $ CRLF $ /' ）。或者，也许有这样做的更好的办法？

It seems this should be possible by piping the input through a few simple sed programs, but I'm not sure how to refer to CRLF in sed (to use in sed 's/<CRLF><CRLF>/$CRLF$/'). Or maybe there's a better way of doing this?

推荐答案

您可以使用SED装饰用{CRLF}月底所有行：

You can use sed to decorate all rows with a {CRLF} at end:

sed 's/$/<CRLF>/'

然后删除所有\\ r \\ n与TR

then remove all \r\n with tr

| tr -d "\r\n"

然后更换双CRLF与\\ n

and then replace double CRLF's with \n

| sed 's/<CRLF><CRLF>/\n/g'

和去除剩余的CRLF的。

and remove leftover CRLF's.

有是一个单行的sed这做这一切在一个周期内，但我似乎现在不能找到它。

There was an one-liner sed which did all this in a single cycle, but I can't seem to find it now.

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