bash Grep 做一件事，然后从下面的行中剪下
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Grep for one thing then cut from the line below
提问by bloodstorm17
Okay so I can't figure this out. Like a file that I using grep in order to get certain information has it set-up like this.
好吧，所以我无法弄清楚这一点。就像我使用 grep 为了获取某些信息而设置的文件一样。
#aaaa
<numbers 123456>
blah
blah
blah
#bbbb
<numbers 2156488>
blah
blah
blah
#cccc
<numbers 5478624>
blah
blah
blah
What I am doing is I am greping for aaaaor bbbbor ccccand the information that i really need is the numbers. As in when i grep for aaaathe thing i want to obtain is really just the numbers right below it. In this case it would 123456
我正在做的是我正在寻找aaaa或bbbb或cccc我真正需要的信息是数字。就像当我搜索aaaa我想要获得的东西时一样，它实际上只是它下面的数字。在这种情况下，它会123456
I know how to grep for aaaabut i don't know how to go to the next line and cut the number.
我知道如何 grep foraaaa但我不知道如何转到下一行并减少数字。
回答by perreal
Using sed:
使用 sed：
sed -n '/aaaa/{
N
s/[^0-9]*\([0-9]*\).*//p
}' input_file
回答by Mihai Maruseac
You can use the -Coption of grepto show one line of context. Then you can tailon the last line (-n 1), cutusing spaces and selecting the second field, cutagain using >and selecting the first field. Thus:
您可以使用-C选项grep来显示一行上下文。然后您可以tail在最后一行 ( -n 1) 上cut使用空格并选择第二个字段，cut再次使用>并选择第一个字段。因此：
$ grep aaaa file -C 1 | tail -n 1 | cut -f2 -d' ' | cut -d'>' -f1
123456
will give back the number you're requesting.
将返回您请求的号码。
The most important part is the -Coption of grep
最重要的部分是-C选项grep
回答by Jonathan Leffler
It looks more like a job for sedto me:
sed对我来说，它看起来更像是一份工作：
sed -n '/^#[a-z]\{4\}/{ N; s/#.*\n<numbers //; s/>//p; }'
The -nsays don't print by default. The /^#[a-z]\{4\}/looks for lines like #aaaa; the actions inside { ... }apply only to such lines. The Nmeans 'read the next line'; the first s///removes the material before the number; the second removes the trailing >and prints.
该-n说默认情况下不打印。在/^#[a-z]\{4\}/看起来像线#aaaa; 里面的动作{ ... }只适用于这些行。的N装置“读取下一行”; 第一个s///去除数字之前的材料；第二个删除尾随>和打印。
There are other options on how to do the 'delete irrelevant material', such as just one substitute command: s/[^0-9]*\([0-9][0-9]+\).*/\1/; this captures the first string of digits and removes everything else. It will handle more variations in the input than the more constrained regular expressions originally shown.
关于如何执行“删除无关材料”还有其他选项，例如只有一个替代命令：s/[^0-9]*\([0-9][0-9]+\).*/\1/; 这将捕获第一串数字并删除其他所有内容。与最初显示的更受约束的正则表达式相比，它将处理更多的输入变化。
Output from sample data:
样本数据的输出：
123456
2156488
5478624
This isn't automatically a job for sed; change it so that the interesting information was the third line after the match and it would be getting fiddly in sed(though N;N;N;probably does what's wanted).
这不是自动的工作sed；更改它，以便有趣的信息出现在比赛后的第三行，并且它会变得繁琐sed（尽管N;N;N;可能会做想要的事情）。
回答by bloodstorm17
@ Lars Kotthoff
@拉尔斯·科特霍夫
Your suggestion to use the -A 1option worked perfectly!
您使用该-A 1选项的建议非常有效！
The answer using your input is this:
使用您的输入的答案是这样的：
grep "aaaa" file -A 1 | grep "<numbers" | cut -d" " -f2 | cut -d">" -f1
I would love to give you the credit for this one!
我很乐意给你这个荣誉！
回答by slitvinov
If you are not insisting on grep+0to get rid of ">"
如果你不是坚持grep+0要去掉“>”
awk 'f{print +0; exit} /^#aaaa/{f=1}' foo.txt
Or
或者
awk 'f{print +0; f=0} /^#aaaa/{f=1}' foo.txt
回答by askmish
Try this one:
试试这个：
for i in "aaaa bbbb"
do
sed -n '/'"$i"'/,+1 p' test |tail -n1|cut -d' ' -f2| sed 's/.$//'
done
Its not very efficient but does the job fine.
它的效率不是很高，但可以很好地完成工作。
