如何通过awk和grep“过滤”尾部输出？

Question

在特殊的控制台上，我喜欢从/ var / log / syslog中过滤一些信息。那不是很棘手：

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";}'

但是现在我想通过grep管道这个特殊字段。只添加一个“| grep Fieldname”不起作用，即使不是grep first，awk稍后（这会更有意义）。

你能给我一个提示吗？

Answer 1

请勿使用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";}'

如果确实需要使用grep，可以使用--line-buffered选项，这样就不会缓冲输出。

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";}'

如果您希望grep输出awk，则应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