仅删除特定的行新行/换行符 [英] Remove new line / line break characters only for specific lines

问题描述

我有以下的输出：

02:01:31
OFFLINE
02:02:31
ONLINE

和我想它变成了：

02:01:31 OFFLINE
02:02:31 ONLINE

我发现删除所有换行/线的方式与打破sed的'：一个; N; $ BA; S / \\ n / / g的！然而它给我接下来这是不是真的，我追求的：

I found a way of removing all newline / line breaks with sed ':a;N;$!ba;s/\n/ /g' however it gives me what follows which is not really what I'm after:

02:01:31 OFFLINE 02:02:31 ONLINE

问：我如何删除换行符/换行符只匹配某种模式（在这种情况下该行 [0-9] 就足够了？

Q: How can I remove newline / line breaks only for lines that match a certain pattern (in this case [0-9] would be enough?

PS：这并不一定与 SED 只要我可以申请一个正则表达式的行的匹配

PS: It doesn't have to be with sed as long as I can apply a regex pattern to the matching of lines

推荐答案

请您 S / \\ n / / G 是 S / \\（ [0-9] \\ S * \\）\\ N / \\ 1 / G （数字后面加一个换行符非空白的任何金额）。

Make your s/\n/ /g be s/\([0-9]\S*\)\n/\1 /g (a digit followed by any amount of non-whitespace followed by a newline).

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