如何从wget输出grep下载速度？

Question

我需要下载几个文件wget并测量下载速度.

我下载了

wget -O /dev/null http://ftp.bit.nl/pub/OpenBSD/4.7/i386/floppy47.fs http://ftp.bit.nl/pub/OpenBSD/4.7/i386/floppyB47.fs

而输出是

--2010-10-11 18:56:00--  http://ftp.bit.nl/pub/OpenBSD/4.7/i386/floppy47.fs
Resolving ftp.bit.nl... 213.136.12.213, 2001:7b8:3:37:20e:cff:fe4d:69ac
Connecting to ftp.bit.nl|213.136.12.213|:80... connected.
HTTP request sent, awaiting response... 200 OK
Length: 1474560 (1.4M) [text/plain]
Saving to: `/dev/null'

100%[==============================================================>] 1,474,560    481K/s   in 3.0s

2010-10-11 18:56:03 (481 KB/s) - `/dev/null' saved [1474560/1474560]

--2010-10-11 18:56:03--  http://ftp.bit.nl/pub/OpenBSD/4.7/i386/floppyB47.fs
Reusing existing connection to ftp.bit.nl:80.
HTTP request sent, awaiting response... 200 OK
Length: 1474560 (1.4M) [text/plain]
Saving to: `/dev/null'

100%[==============================================================>] 1,474,560    499K/s   in 2.9s

2010-10-11 18:56:06 (499 KB/s) - `/dev/null' saved [1474560/1474560]

FINISHED --2010-10-11 18:56:06--
Downloaded: 2 files, 2.8M in 5.9s (490 KB/s)

我需要grep总下载速度,即字符串490 KB/s.我该怎么做呢？

PS可能需要考虑我们实际上只下载一个文件的情况,因此不会有最终输出 FINISHED

Answer 1

更新，使用 sed 的 grep 风格版本：

wget ... 2>&1 | sed -n '$,$s/.*(\(.*\)).*/\1/p'

旧版本：

我想，下载后用文件大小除以下载时间更容易。;-)

(/usr/bin/time -p wget ... 2>&1 >/dev/null; ls -l newfile) | \
awk '
   NR==1 {t=$2};
   NR==4 {printf("rate=%f bytes/second\n", $5/t)}
'

awk 的第一行将“real xx.xx”的实际经过时间存储在 variabe 中t。第二条 awk 行将文件大小（ 的第 5 列ls -l）除以时间并将其输出为速率。