使用grep或awk

Question

我在日志文件中有一行看起来像

Oct 07, 2014 7:39:10 AM x.y.z

SEVERE: adding the post (STORY)  abcd = 495274900579805_10204277254604731 : a = 0 b = 0 c = 0

我想从第一行得到日期和时间，从第二行得到a = 0 b = 0 c = 0，我怎么能用grep和awk实现这个目的。请帮助

Answer 1

你可以尝试下面的grep命令，

$ grep -oP '.*?(?=\s[^.\s]+\.[^.\s]+\.\S+)|:\s+\K[^:]*$' file
Oct 07, 2014 7:39:10 AM
a = 0 b = 0 c = 0

<强>更新

$ grep -oP '.*?(?=\s[^.\s]+\.[^.\s]+\.\S+)|\) *\K.*' file
Oct 07, 2014 7:39:10 AM
abcd = 495274900579805_10204277254604731 : a = 0 b = 0 c = 0

Answer 2

这是awk版本（您要求使用awk）

awk '/AM|PM/ && NF--; /a =/ {print "a = "$(NF-6),"b = "$(NF-3),"c = "$NF}' file
Oct 07, 2014 7:39:10 AM
a = 0 b = 0 c = 0

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另一个版本：

awk '/AM|PM/ && NF--; {n=split($0,a,"abcd");if (n==2) print "abcd"a[2]}' file
Oct 07, 2014 7:39:10 AM
abcd = 495274900579805_10204277254604731 : a = 0 b = 0 c = 0