bash sed/awk：从文本流中提取模式
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sed/awk: Extract pattern from text stream
提问by Dagang
2011-07-01 ... /home/todd/logs/server_log_1.log ...
2011-07-02 ... /home/todd/logs/server_log_2.log ...
2011-07-03 ... /home/todd/logs/server_log_3.log ...
I have a file looks like the above. I want to extract the file names from it and output to STDOUT as:
我有一个看起来像上面的文件。我想从中提取文件名并输出到 STDOUT 为：
server_log_1.log
server_log_2.log
server_log_3.log
Could someone help? Thanks!
有人可以帮忙吗？谢谢！
The file name pattern is server_log_xxx.log, and it only occurs once in a line.
文件名模式是 server_log_xxx.log，它在一行中只出现一次。
回答by glenn Hymanman
Assuming the "xxx" placeholder is only digits:
假设“xxx”占位符只是数字：
grep -o 'server_log_[0-9]\+\.log'
回答by Pawe? Nadolski
Pipe your file through following command:
通过以下命令管道您的文件：
sed 's/.*\(server_log_[0-9]\+\.log\).*//'
回答by Zsolt Botykai
With awk and your input pattern:
使用 awk 和您的输入模式：
awk 'BEGIN {FS="/"}
{ print gensub(" .*$","","g",) }' INPUTFILE
See it action here: https://ideone.com/kcadh
在此处查看操作：https: //ideone.com/kcadh
HTH
HTH
回答by Dimitre Radoulov
sed 's|.*/\([^/ ]*\).*||' infile
