regex - 来自终端的 Sed 正则表达式字符串替换

Question

我有一个标准格式的日志文件，例如:

31 Mar - Lorem Ipsom1
31 Mar - Lorem Ipsom2
31 Mar - Lorem Ipsom3

我要实现的替换是 31*31 到 31，所以我最终会得到一个只有最后一行的日志，在这个例子中它看起来像:

31 Mar - Lorem Ipsom3

我希望在没有 perl 的定制 linux 机器上执行它。 我试过像这样使用 sed:

sed -i -- 's/31*31/31/g' /var/log/prog/logFile

但它什么也没做.. 也欢迎任何涉及忍者 bash 命令的替代方法。

Answer 1

一种只保留最后一行匹配模式的方法是

sed -n '/^31/ { :a $!{ h; n; //ba; x; G } }; p' filename

工作原理如下:

/^31/ {    # if a line begins with 31
  :a       # jump label for looping

  $!{      # if the end of input has not been reached (otherwise the current
           # line is the last line of the block by virtue of being the last
           # line)

    h      # hold the current line
    n      # fetch the next line. (note that this doesn't print the line
           # because of -n)

    //ba   # if that line also begins with 31, go to :a. // attempts the
           # most recently attempted regex again, which was ^31

    x      # swap hold buffer, pattern space
    G      # append hold buffer to pattern space. The PS now contains
           # the last line of the block followed by the first line that 
           # comes after it
  }
}
p          # in the end, print the result

这避免了多行正则表达式的一些问题，例如在一行中间开始或结束的匹配。它也不会丢弃两个匹配行 block 之间的行，并保留每个 block 的最后一行。