awk:拆分为"\ n" [英] awk: Split on "\n"

问题描述

我正在尝试处理一个日志文件，其中的条目被压缩为一行，换行符编码为"\ n".我想将所有内容都保留到第一个"\ n"，其余的都丢弃. awk -F"\ n"'{print $ 1}'文件不起作用， awk -F"\\ n"'{print $ 1}'文件.该命令的正确形式是什么?

I'm trying to process a log file in which entries are compressed into one line with the newline encoded as "\n". I want to keep everything up to the first "\n" and discard the rest. awk -F"\n" '{print $1}' file doesn't work, and neither does awk -F"\\n" '{print $1}' file. What's the correct form of this command?

推荐答案

$ echo 'a\nb'
a\nb

$ echo 'a\nb' | awk -F'\\\\n' '{print $1}'
a

原因:在正则表达式比较中考虑上述字符的这些用法:

Here's why: Consider these uses of the above characters in regexp comparisons:

• n =文字字符 n ( $ 0〜/n/)
• \ n =文字换行符( $ 0〜/\ n/)
• \\ =在正则表达式常量( $ 0〜/\\/)中使用反斜杠
• \\\\ =在动态正则表达式中使用反斜杠( $ 0〜"\\\\" )

• n = the literal character n ($0 ~ /n/)
• \n = a literal newline character ($0 ~ /\n/)
• \\ = a backslash when used in a regexp constant ($0 ~ /\\/)
• \\\\ = a backslash when used in a dynamic regexp ($0 ~ "\\\\")

最后一个是因为动态正则表达式是一个字符串，必须将其解析一次才能转换为正则表达式，然后在用作该正则表达式时再次进行解析，因此，由于它被解析了两次，因此需要将所有转义符加倍

That last one is because a dynamic regexp is a string which has to be parsed once to be converted to a regexp and then gets parsed again when used as that regexp, so since it gets parsed twice it needs all escapes to be doubled.

由于当您说 -F无论如何" 时，字段分隔符基本上是一个正则表达式(有一些曲折)，因此您将FS变量定义为动态正则表达式，因此转义必须加倍

Since a field separator is basically a regexp (with a few twists) when you say -F "whatever" you are defining the FS variable to be a dynamic regexp and so escapes have to be doubled.

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