bash 我可以只 grep 文件的前 n 行吗？
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Can I grep only the first n lines of a file?
提问by David LeBauer
I have very long log files, is it possible to ask grep to only search the first 10 lines?
我有很长的日志文件，是否可以要求 grep 只搜索前 10 行？
回答by Joachim Isaksson
The magic of pipes;
管道的魔力；
head -10 log.txt | grep <whatever>
回答by cxw
For folks who find this on Google, I needed to search the first nlines of multiple files, but to only print the matching filenames. I used
对于在谷歌上找到这个的人，我需要搜索n多个文件的第一行，但只打印匹配的文件名。我用了
gawk 'FNR>10 {nextfile} /pattern/ { print FILENAME ; nextfile }' filenames
The FNR..nextfilestops processing a file once 10 lines have been seen. The //..{}prints the filename and moves on whenever the first match in a given file shows up. To quote the filenames for the benefit of other programs, use
FNR..nextfile一旦看到 10 行，就会停止处理文件。该//..{}上打印的文件名和移动，每当在给定的文件显示了第一场比赛。要引用文件名以使其他程序受益，请使用
gawk 'FNR>10 {nextfile} /pattern/ { print "\"" FILENAME "\"" ; nextfile }' filenames
回答by Zsolt Botykai
Or use awkfor a single process without |:
或awk用于单个进程而没有|：
awk '/your_regexp/ && NR < 11' INPUTFILE
On each line, if your_regexpmatches, and the number of records (lines) is less than 11, it executes the default action (which is printing the input line).
在每一行上，如果your_regexp匹配，并且记录（行）数小于 11，则执行默认操作（即打印输入行）。
Or use sed:
或使用sed：
sed -n '/your_regexp/p;10q' INPUTFILE
Checks your regexp and prints the line (-nmeans don't print the input, which is otherwise the default), and quits right after the 10th line.
检查您的正则表达式并打印该行（-n意味着不打印输入，否则为默认值），并在第 10 行之后立即退出。
回答by Dan Fego
You have a few options using programs along with grep. The simplest in my opinion is to use head:
您有几个选项可以使用程序和grep. 我认为最简单的方法是使用head：
head -n10 filename | grep ...
headwill output the first 10 lines (using the -noption), and then you can pipe that output to grep.
head将输出前 10 行（使用-n选项），然后您可以将该输出通过管道传输到grep.
回答by jaypal singh
grep "pattern" <(head -n 10 filename)
回答by Dileepa Chandima
grep -m6 "string" cov.txt
This searches only the first 6 lines for string
这仅搜索前 6 行 string
回答by vins
head -10 log.txt | grep -A 2 -B 2 pattern_to_search
-A 2: print two lines before the pattern.
-A 2: 在图案前打印两行。
-B 2: print two lines after the pattern.
-B 2：在图案后打印两行。
head -10 log.txt # read the first 10 lines of the file.
回答by Alan Haggai Alavi
The output of head -10 filecan be piped to grepin order to accomplish this:
的输出head -10 file可以通过管道传输grep以完成此操作：
head -10 file | grep …
Using Perl:
使用 Perl：
perl -ne 'last if $. > 10; print if /pattern/' file
回答by Gustavo Straube
You can use the following line:
您可以使用以下行：
head -n 10 /path/to/file | grep [...]
回答by RoG
An extension to Joachim Isaksson's answer: Quite often I need something from the middle of a long file, e.g. lines 5001 to 5020, in which case you can combine headwith tail:
Joachim Isaksson 答案的扩展：我经常需要长文件中间的一些内容，例如第 5001 行到 5020 行，在这种情况下，您可以结合head使用tail：
head -5020 file.txt | tail -20 | grep x
This gets the first 5020 lines, then shows only the last 20 of those, then pipes everything to grep.
这将获取前 5020 行，然后仅显示其中的最后 20 行，然后将所有内容通过管道传送到 grep。
(Edited: fencepost error in my example numbers, added pipe to grep)
（编辑：我的示例数字中的围栏错误，向 grep 添加了管道）
