perl - 捕获独特的部分，无论出现偶数还是奇数

Question

我有一个文本文件，其中反复出现一组特定的连续行。我需要修剪所有重复出现的情况，只打印第一个出现的情况。

输入:

$ cat log_repeat.txt
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554

$

下面的 Perl 解决方案仅在出现奇数情况时才有效，

$ perl -0777 -pe 's/(^total.*)\1//gms ' log_repeat.txt
total bytes = 0, at time = 1190554
time window = 0, at time = 1190554
BW in Mbps = 0, at time = 1190554

$

并且当出现偶数时不打印任何内容。无论该部分重复奇数次还是偶数次，如何获得第一次出现的结果。

Answer 1

贪婪地多次匹配你的 block ，只要后面跟着另一个 block

perl -0777 -wpe's/(total.*)+(?=\1)//s' log_repeat.txt

前瞻确保保留一个(最后一个)，因为它不会消耗其匹配项。

或者，保留第一个匹配项，用 \K 丢弃它，并删除其他匹配项

perl -0777 -wpe's/(total.*?)\K\1+//s' log_repeat.txt

请注意，此处必须使用的 .*? 与 .* 有所不同，但可能不实用。