从iperf结果中提取最高值

Question

我需要从iperf日志文件输出中提取最高值，忽略第7行中的第一个条目。

------------------------------------------------------------
Client connecting to 10.10.10.2, TCP port 5001
TCP window size: 0.02 MByte (default)
------------------------------------------------------------
[  3] local 10.10.10.1 port 44809 connected with 10.10.10.2 port 5001
[ ID] Interval       Transfer     Bandwidth
[  3]  0.0- 1.0 sec  2.50 MBytes  21.0 Mbits/sec
[  3]  1.0- 2.0 sec  2.25 MBytes  18.9 Mbits/sec
[  3]  2.0- 3.0 sec  2.00 MBytes  16.8 Mbits/sec
[  3]  3.0- 4.0 sec  2.25 MBytes  18.9 Mbits/sec
[  3]  4.0- 5.0 sec  2.00 MBytes  16.8 Mbits/sec
[  3]  5.0- 6.0 sec  2.25 MBytes  18.9 Mbits/sec
[  3]  6.0- 7.0 sec  2.12 MBytes  17.8 Mbits/sec
[  3]  7.0- 8.0 sec  2.00 MBytes  16.8 Mbits/sec
[  3]  8.0- 9.0 sec  2.25 MBytes  18.9 Mbits/sec
[  3]  9.0-10.0 sec  2.00 MBytes  16.8 Mbits/sec
[  3]  0.0-10.1 sec  21.8 MBytes  18.0 Mbits/sec

到目前为止我已经

了

#!/bin/bash
iperf -c 10.0.0.2 -t 10 -f m -i 1 > /tmp/max.log
sed -i '7d' /tmp/max.log

我现在需要回显结果列18.9中的最高值

Answer 1

您可以不使用sed，只需告诉awk忽略第7行的所有内容。

$ awk 'NR > 7 { max = $(NF-1) > max ? $(NF-1) : max } END { print max }' m.txt
18.9