[ACCEPTED]-How to output lines 800-900 of a file with a unix command?-head

Accepted answer
Score: 26
sed -n '800,900p' file.txt

This will print (p) lines 800 through 900, including 6 both line 800 and 900 (i.e. 101 lines in 5 total). It will not print any other lines 4 (-n).

Adjust from 800 to 801 and/or 900 to 3 899 to make it do exactly what you think 2 "between 800 and 900" should mean in your 1 case.

Score: 5

Found a prettier way: Using sed, to print 5 out only lines between a and b:

sed -n -e 800,900p filename.txt

From the 4 blog post: Using sed to extract lines in a text file

One way I am using it is to 3 find (and diff) similar sections of files:

 sed -n -e 705,830p mnetframe.css > tmp1; \
 sed -n -e 830,955p mnetframe.css > tmp2; \
 diff --side-by-side tmp1 tmp2

Which 2 will give me a nice side-by-side comparison 1 of similar sections of a file :)

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