sed，替换#include中的反斜杠

Question

I wish to replace all the backslashes (which appear on the same line with an include directive) with slashes.

Here's what I have until now..

echo '#include "..\etc\filename\yes"' | sed 's&\(#include.*\)\\&\1\/&g'

This works as I expect, but the problem is that it replaces only one \\ at a time... If I want to replace all three in the above text, I have to run the sed command 3 times... The g flag at the end should make the replacements globally, no?

I'm using sed 4.2.1 on Ubuntu 11.10...

Answer 1

The problem is the way you're matching. The .* is greedy, so it matches the last backslash first and then thinks it's done. Try this:

... | sed '/^#include/s&\\&/&g'

That runs the substitutions only on lines matching the first pattern.

Answer 2

你想要一个复合命令 - 第一个模式匹配以#include开头的行，第二个模式匹配你的斜杠翻译。

sed '/^#include/ s&\\&/&g'