我有一个~23000行的SQL转储，其中包含几个数据库的数据价值。我需要提取这个文件的某个部分(即单个数据库的数据)，并将其放在一个新文件中。我知道我想要的数据的开始行号和结束行号。
谁知道一个Unix命令(或一系列命令)可以从文件中提取16224到16482行之间的所有行，然后将它们重定向到一个新文件中?
当前回答
使用ruby:
ruby -ne 'puts "#{$.}: #{$_}" if $. >= 32613500 && $. <= 32614500' < GND.rdf > GND.extract.rdf
其他回答
我已经为sed、perl、head+tail和我自己的awk代码编译了一些最高评级的解决方案，并通过管道关注性能，同时使用LC_ALL=C确保所有候选程序以尽可能快的速度运行，并在两者之间分配2秒的睡眠间隔。
差距是显而易见的:
abs time awk/app speed ratio
----------------------------------
0.0672 sec : 1.00x mawk-2
0.0839 sec : 1.25x gnu-sed
0.1289 sec : 1.92x perl
0.2151 sec : 3.20x gnu-head+tail
还没有机会测试这些工具的python或BSD变体。
(fg && fg && fg && fg) 2>/dev/null;
echo;
( time ( pvE0 < "${m3t}"
| LC_ALL=C mawk2 '
BEGIN {
_=10420001-(\
__=10420256)^(FS="^$")
} _<NR {
print
if(__==NR) { exit }
}' ) | pvE9) | tee >(xxh128sum >&2) | LC_ALL=C gwc -lcm | lgp3 ;
sleep 2;
(fg && fg && fg && fg) 2>/dev/null
echo;
( time ( pvE0 < "${m3t}"
| LC_ALL=C gsed -n '10420001,10420256p;10420256q'
) | pvE9 ) | tee >(xxh128sum >&2) | LC_ALL=C gwc -lcm | lgp3 ;
sleep 2; (fg && fg && fg && fg) 2>/dev/null
echo
( time ( pvE0 < "${m3t}"
| LC_ALL=C perl -ne 'print if 10420001..10420256'
) | pvE9 ) | tee >(xxh128sum >&2) | LC_ALL=C gwc -lcm | lgp3 ;
sleep 2; (fg && fg && fg && fg) 2>/dev/null
echo
( time ( pvE0 < "${m3t}"
| LC_ALL=C ghead -n +10420256
| LC_ALL=C gtail -n +10420001
) | pvE9 ) | tee >(xxh128sum >&2) | LC_ALL=C gwc -lcm | lgp3 ;
in0: 1.51GiB 0:00:00 [2.31GiB/s] [2.31GiB/s] [============> ] 81%
out9: 42.5KiB 0:00:00 [64.9KiB/s] [64.9KiB/s] [ <=> ]
( pvE 0.1 in0 < "${m3t}" | LC_ALL=C mawk2 ; )
0.43s user 0.36s system 117% cpu 0.672 total
256 43487 43487
54313365c2e66a48dc1dc33595716cc8 stdin
out9: 42.5KiB 0:00:00 [51.7KiB/s] [51.7KiB/s] [ <=> ]
in0: 1.51GiB 0:00:00 [1.84GiB/s] [1.84GiB/s] [==========> ] 81%
( pvE 0.1 in0 < "${m3t}" |LC_ALL=C gsed -n '10420001,10420256p;10420256q'; )
0.68s user 0.34s system 121% cpu 0.839 total
256 43487 43487
54313365c2e66a48dc1dc33595716cc8 stdin
in0: 1.85GiB 0:00:01 [1.46GiB/s] [1.46GiB/s] [=============>] 100%
out9: 42.5KiB 0:00:01 [33.5KiB/s] [33.5KiB/s] [ <=> ]
( pvE 0.1 in0 < "${m3t}" | LC_ALL=C perl -ne 'print if 10420001..10420256'; )
1.10s user 0.44s system 119% cpu 1.289 total
256 43487 43487
54313365c2e66a48dc1dc33595716cc8 stdin
in0: 1.51GiB 0:00:02 [ 728MiB/s] [ 728MiB/s] [=============> ] 81%
out9: 42.5KiB 0:00:02 [19.9KiB/s] [19.9KiB/s] [ <=> ]
( pvE 0.1 in0 < "${m3t}"
| LC_ALL=C ghead -n +10420256
| LC_ALL=C gtail -n ; )
1.98s user 1.40s system 157% cpu 2.151 total
256 43487 43487
54313365c2e66a48dc1dc33595716cc8 stdin
那些想要计算头部、|、尾部组合的间隔的人是想多了。
下面是如何在不计算任何东西的情况下得到“16224 - 16482”范围:
cat file | head -n +16482 | tail -n +16224
解释:
The + instructs the head/tail command to "go up to / start from" (respectively) the specified line number as counted from the beginning of the file. Similarly, a - instructs them to "go up to / start from" (respectively) the specified line number as counted from the end of the file The solution shown above simply uses head first, to 'keep everything up to the top number', and then tail second, to 'keep everything from the bottom number upwards', thus defining our range of interest (with no need to compute an interval).
我想从一个使用变量的脚本中做同样的事情，并通过在$变量周围加上引号来分隔变量名和p来实现:
sed -n "$first","$count"p imagelist.txt >"$imageblock"
我想把一个列表分成不同的文件夹，找到最初的问题和答案，这是一个有用的步骤。(分裂命令不是旧操作系统上的选项，我必须将代码移植到)。
sed -n '16224,16482p;16483q' filename > newfile
来自sed手册:
p - Print out the pattern space (to the standard output). This command is usually only used in conjunction with the -n command-line option. n - If auto-print is not disabled, print the pattern space, then, regardless, replace the pattern space with the next line of input. If there is no more input then sed exits without processing any more commands. q - Exit sed without processing any more commands or input. Note that the current pattern space is printed if auto-print is not disabled with the -n option.
and
sed脚本中的地址可以是以下任何一种形式: 数量 指定行号将只匹配输入中的该行。 一个地址范围可以通过指定两个地址来指定 用逗号(，)分隔。地址范围匹配从 第一个地址匹配，并一直持续到第二个 地址匹配(包括)。
perl -ne 'print if 16224..16482' file.txt > new_file.txt
