Linux命令行帮助| 从文件中提取特定部分

Question

Suppose I have a log which has data in the format given below

        Time           number status  
2013-5-10 19:18:43.430 123456 success 
2013-5-10 19:28:13.430 134324 fail 
2013-5-10 19:58:33.430 456456 success 

I want to extract the numbers having success status. Is there any way in linux using command line(grep, sed) to extract the data as mentioned. ?? Thanks all ..

Answer 1

grep only solution:

grep -Po '\d+(?= success)' file

or with awk only :

awk '$4=="success"&&$0=$3' input

Answer 2

cat file | grep success | awk '{print $3}'

Answer 3

你可以做

(grep 'success' | cut -d ' ' -f 3) <$file

Answer 4

这将根据成功状态打印数字：

 awk '$4 ~ /success/ {print $3}' logfile

Answer 5

使用perl：

perl -ne '/success/ && split && print "$_[2]\n"' inputfile