Bash 匹配多行字符串并附加到上一行

Question

我正在尝试用这些行转换文本文件：

policy rule {
    source-address host1;
    destination-address host2;
    application http;
policy someotherrule {
    source-address any;
    destination-address [ host2 host3];
    application ssh;

进入：

policy rule { source-address host1; destination-address host2; application http;
policy someotherrule { source-address any; destination-address [ host2 host3]; application ssh;

所以基本上，如果一行以分号结尾，将其附加到上一行。

我一直在尝试 sed/awk，但还没有找到一个有效的解决方案来获得所需的输出。

Answer 1

以下 gnu-awk 应该可以工作：

awk -v RS='(^|\n)[^\n]*[^;]\n' 'NF{s=p [=10=]; gsub(/(^|\n+)[[:blank:]]+/, " ", s); print s}
          {p=RT}' file
 policy rule { source-address host1; destination-address host2; application http;
 policy someotherrule { source-address any; destination-address [ host2 host3]; application ssh;

Answer 2

这条简单的单行线就可以做到：

perl -p0e 's/\n\s+/ /g' file
policy rule { source-address host1; destination-address host2; application http;
policy someotherrule { source-address any; destination-address [ host2 host3]; application ssh;

Answer 3

这是一个awk解决方案：

awk '{=;printf (/^policy/ &&NR>1?RS:"")"%s",[=10=]} END {print ""}' file
policy rule {source-address host1;destination-address host2;application http;
policy someotherrule {source-address any;destination-address [ host2 host3];application ssh;

它根据行是否以 policy 开头来设置行输出的变化。

Bash 匹配多行字符串并附加到上一行

Bash match string in multiple lines and append to previous line

linux

bash

awk

sed