awk |在字段匹配的基础上合并行 [英] awk | merge line on the basis of field matching

问题描述

我需要以下方面的帮助:

I need help with following:

输入文件:

abc message=sent session:111,x,y,z
pqr message=receive session:111,4,5,7
abc message=sent session:123,x,y,z
pqr message=receive session:123,4,5,7
abc message=sent session:342,x,y,z
abc message=sent session:589,x,y,z
pqr message=receive session:589,4,5,7

输出文件:

abc message=sent session:111,x,y,z, pqr message=receive session:111,4,5,7
abc message=sent session:123,x,y,z, pqr message=receive session:123,4,5,7
abc message=sent session:342,x,y,z, NOMATCH
abc message=sent session:589,x,y,z, pqr message=receive session:589,4,5,7

注意事项:

如果你在源文件中看到，对于每个发送"的消息，都有一个接收"
只有 session=342 没有接收
会话未知，不能硬编码
所以只合并那些我们有匹配会话号的发送和接收

If you see in source file, for every "sent" message there is "receive"
only for session=342 there is no receive
session is unknow, can't be hardcoded
So merge only those sent and receive where we have matching session number

推荐答案

另一种方式:

awk -F "[:,]"  '/=sent/{a[$2]=$0;}/=receive/{print a[$2], $0;delete a[$2];}END{for(i in a)print a[i],"NO MATCH";}' file

结果:

abc message=sent session:111,x,y,z pqr message=receive session:111,4,5,7
abc message=sent session:123,x,y,z pqr message=receive session:123,4,5,7
abc message=sent session:589,x,y,z pqr message=receive session:589,4,5,7
abc message=sent session:342,x,y,z NO MATCH

当遇到send记录时，将其存储在以会话ID为索引的数组中.当遇到receive 记录时，从数组中取出send 记录并与receive 记录一起打印.此外，当接收到 receive 记录时，发送的记录将从数组中删除.在 END 处，数组中的所有剩余记录都打印为 NO MATCH.

When the send record is encountered, it is store in the array with the session id as the index. When the receive record is encountered, the send record is fetched from the array and printed along with receive record. Also, sent records are removed from array as and when receive records are received. At the END, all the remaining records in the array are printed as NO MATCH.

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