如何通过 awk 和 grep “过滤”尾部输出？

Question

在一个特殊的控制台上，我喜欢从 /var/log/syslog 中过滤一些信息。这并不是很棘手：

tail -f /var/log/syslog | awk '{print $2,$1,$9,$3,"\033[1;36m"$17 "\033[0m","\033[1;33m"$23 "\033[0m","\033[1;36m"$19 "\033[0m","\033[1;33m"$24 "\033[0m","\033[1;38m"$26"\033[0m","\033[1;32m"$13"\033[0m","\033[1;31m"$20 "\033[0m";}'

但现在我想通过 grep 将其用于特殊字段。仅仅添加一个“| grep Fieldname”是行不通的，甚至不是先grep，然后awk（这更有意义）。

你能给我小费吗？

Answer 1

不要使用grep，在awk中进行模式匹配。

tail -f /var/log/syslog | awk '/Fieldname/ {print $2,$1,$9,$3,"\033[1;36m"$17 "\033[0m","\033[1;33m"$23 "\033[0m","\033[1;36m"$19 "\033[0m","\033[1;33m"$24 "\033[0m","\033[1;38m"$26"\033[0m","\033[1;32m"$13"\033[0m","\033[1;31m"$20 "\033[0m";}'

如果您确实需要使用grep，可以使用--line-buffered 选项，这样它就不会缓冲输出。

tail -f /var/log/syslog | grep --line-buffered Fieldname | awk '{print $2,$1,$9,$3,"\033[1;36m"$17 "\033[0m","\033[1;33m"$23 "\033[0m","\033[1;36m"$19 "\033[0m","\033[1;33m"$24 "\033[0m","\033[1;38m"$26"\033[0m","\033[1;32m"$13"\033[0m","\033[1;31m"$20 "\033[0m";}'

如果你想grepawk的输出，你应该在打印每一行后使用fflush()立即刷新缓冲区。

tail -f /var/log/syslog | awk '{print $2,$1,$9,$3,"\033[1;36m"$17 "\033[0m","\033[1;33m"$23 "\033[0m","\033[1;36m"$19 "\033[0m","\033[1;33m"$24 "\033[0m","\033[1;38m"$26"\033[0m","\033[1;32m"$13"\033[0m","\033[1;31m"$20 "\033[0m"; fflush();}' | grep Fieldname