Quick unix command to display specific lines in the middle of a file?

Trying to debug an issue with a server and my only log file is a 20GB log file (with no timestamps even! Why do people use System.out.println() as logging? In production?!)

Using grep, I've found an area of the file that I'd like to take a look at, line 347340107.

Other than doing something like

head -<$LINENUM + 10> filename | tail -20 

... which would require head to read through the first 347 million lines of the log file, is there a quick and easy command that would dump lines 347340100 - 347340200 (for example) to the console?

update I totally forgot that grep can print the context around a match ... this works well. Thanks!

---

用GNU-grep你可以说

grep --context=10 ...

---

I found two other solutions if you know the line number but nothing else (no grep possible):

Assuming you need lines 20 to 40,

sed -n '20,40p;41q' file_name

or

awk 'FNR>=20 && FNR<=40' file_name

---

# print line number 52
sed -n '52p' # method 1
sed '52!d' # method 2
sed '52q;d' # method 3,  efficient on large files 

method 3 efficient on large files

fastest way to display specific lines

链接地址: http://www.djcxy.com/p/19652.html

上一篇: 如何截断由grep或ack返回的长匹配行

下一篇: 快速unix命令显示文件中间的特定行？