如何僅用正則表達式在Perl中轉換數據？

Question

我想用多個正則表達式將多行數據轉換成一行。

數據包含HTTP交換的開始和結束通知，後面跟着一行或多行的詳細信息。這裏是做這項工作的正則表達式：

(\d+-\d+-\d+) (\d+:\d+:\d+,\d+) INFO (\[com\.LoggingFilter\]) \(([^)]+)\) (\d+) \* Server in-bound request 
(?=\5 > (.+) 
[\s\S]* 
\5 > user-agent:\s*(.+) 
[\s\S]* 
(\d+-\d+-\d+) (\d+:\d+:\d+,\d+) INFO \3 \(\4\) \5 \* Server out-bound response 
\5 < (\d+) 
[\s\S]*) 

它（幾乎）工作原理：VIEW regex in action。在正則表達式替換之後，一些細節（由積極的展望保存）仍然保留。如何得到以下結果的

"88598";"2016-07-29";"00:00:08,262";"2016-07-29";"00:00:08,262";"http-10.2.3.4-8080-14";"Java/1.7.0_79";"303";"GET http://1.2.3.4:8080/service/api/ 
"88599";"2016-07-29";"00:00:08,382";"2016-07-29";"00:00:08,382";"http-10.2.3.4-8080-8";"Java/1.7.0_79";"303";"GET http://1.2.3.4:8080/service/api/" 
(...) 

代替

"88598";"2016-07-29";"00:00:08,262";"2016-07-29";"00:00:08,262";"http-10.2.3.4-8080-14";"Java/1.7.0_79";"303";"GET http://1.2.3.4:8080/service/api/"88598 > GET http://1.2.3.4:8080/service/api/ 
<garbage data here> 
"88599";"2016-07-29";"00:00:08,382";"2016-07-29";"00:00:08,382";"http-10.2.3.4-8080-8";"Java/1.7.0_79";"303";"GET http://1.2.3.4:8080/service/api/"88599 > GET http://1.2.3.4:8080/service/api/ 
(...) 

Answer 1

這似乎工作：

(\d+-\d+-\d+) (\d+:\d+:\d+,\d+) INFO (\[com\.LoggingFilter\]) \(([^)]+)\) (\d+) \* Server in-bound request 
(?=\5 > (.+) 
[\s\S]* 
\5 > user-agent:\s*(.+) 
[\s\S]* 
(\d+-\d+-\d+) (\d+:\d+:\d+,\d+) INFO \3 \(\4\) \5 \* Server out-bound response 
\5 < (\d+) 
[\s\S]*)[\s\S]*?((?=\n.*Server in-bound request)|(?![\s\S])) 

Here演示。