sed awk用模式匹配替换行

Question

使用awk或sed如何在模式匹配后替换每一行？ 该模式总是以S：\和其他方式开始， 我需要从S：\〜到最后的整行出现在空白行之前的下一行。

我有这样的输入：

S:\dir1\subfolder1longsubf
abcdefg
1234567
permissions

S:\dir2\verylongsub
some random string
some random string

S:\dir3
some random string
some random string

S:\dir4\sub2\sub3
some random string
some random string
some random string

我需要这样的输出：

S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf

S:\dir2\verylongsub
S:\dir2\verylongsub
S:\dir2\verylongsub

S:\dir3
S:\dir3
S:\dir3

S:\dir4\sub2\sub3
S:\dir4\sub2\sub3
S:\dir4\sub2\sub3
S:\dir4\sub2\sub3

Answer 1

我会使用awk：

awk '/^S/ {line=$0} {print NF?line:""}' file

这会存储以S开头的行。然后，如果该行为空，则打印该存储的值或空行。

测试

$ awk '/^S/ {line=$0} {print NF?line:""}' file
S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf
S:\dir1\subfolder1longsubf

S:\dir2\verylongsub
S:\dir2\verylongsub
S:\dir2\verylongsub

S:\dir3
S:\dir3
S:\dir3

S:\dir4\sub2\sub3
S:\dir4\sub2\sub3
S:\dir4\sub2\sub3
S:\dir4\sub2\sub3

Answer 2

使用GNU sed：

sed  '/^S:/{h};/^[^S]/{g}' file