Regex 如何仅在记录2与模式匹配时连接记录

我有一个包含时间戳和丢包信息的文件：

Tue May 13 09:50:00 HST 2014
Tue May 13 09:55:00 HST 2014
60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014
60 packets transmitted, 54 packets received, 10.0% packet loss
Tue May 13 10:05:00 HST 2014
Tue May 13 10:10:00 HST 2014
Tue May 13 10:15:00 HST 2014

使用awk，我只想输出数据包信息行，与它前面的日期行相连：

Tue May 13 09:55:00 HST 2014 60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014 60 packets transmitted, 54 packets received, 10.0% packet loss

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我找不到任何关于基于模式匹配加入记录的信息。任何帮助都将不胜感激

您可以使用此

sed

sed -n 'N; /packet/s/\n/ /p; D;' yourfile

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您可以使用此

sed

sed -n 'N; /packet/s/\n/ /p; D;' yourfile

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您可以使用此

sed

sed -n 'N; /packet/s/\n/ /p; D;' yourfile

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您可以使用此

sed

sed -n 'N; /packet/s/\n/ /p; D;' yourfile

使用

awk

awk '/loss$/ {print p,$0} {p=$0}' file
Tue May 13 09:55:00 HST 2014 60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014 60 packets transmitted, 54 packets received, 10.0% packet loss

使用

awk

awk '/loss$/ {print p,$0} {p=$0}' file
Tue May 13 09:55:00 HST 2014 60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014 60 packets transmitted, 54 packets received, 10.0% packet loss

使用

awk

awk '/loss$/ {print p,$0} {p=$0}' file
Tue May 13 09:55:00 HST 2014 60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014 60 packets transmitted, 54 packets received, 10.0% packet loss

使用

awk

awk '/loss$/ {print p,$0} {p=$0}' file
Tue May 13 09:55:00 HST 2014 60 packets transmitted, 50 packets received, 16.7% packet loss
Tue May 13 10:00:00 HST 2014 60 packets transmitted, 54 packets received, 10.0% packet loss

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很好的解决方案！我还没有考虑过sed。很好的解决方案！我还没有考虑过sed。很好的解决方案！我还没有考虑过sed。很好的解决方案！我没有考虑过塞德。