Bash在多行中匹配字符串並追加到前一行

Question

我正在嘗試使用以下行來打開文本文件：

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

進入：

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

因此，基本上，如果一行以分號結尾，請將其附加到前一行。

我一直在與sed / awk玩，但是還沒有找到可行的解決方案來獲得所需的輸出。

Answer 1

以下gnu-awk應該可以工作：

awk -v RS='(^|\n)[^\n]*[^;]\n' 'NF{s=p $0; 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

這個簡單的單行代碼可以做到：

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

這是一個awk解決方案：

awk '{$1=$1;printf (/^policy/ &&NR>1?RS:"")"%s",$0} END {print ""}' file
policy rule {source-address host1;destination-address host2;application http;
policy someotherrule {source-address any;destination-address [ host2 host3];application ssh;

它根據行是否以policy開頭來設置更改行的輸出。