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I need to extract a set number of lines from a file given the start line number and end line number.
How could I quickly do this under unix (it's actually Solaris so gnu flavour isn't available).
Thx
940
Printing given range of lines using sed-n option is to print 'n' number of lines. in above example 'p' is used to Print the current pattern space. Same as above we can also use 'd' Delete pattern space.
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.
The grep command searches through the file, looking for matches to the pattern specified. To use it type grep , then the pattern we're searching for and finally the name of the file (or files) we're searching in. The output is the three lines in the file that contain the letters 'not'.
Use the tail command to write the file specified by the File parameter to standard output beginning at a specified point. This displays the last 10 lines of the accounts file. The tail command continues to display lines as they are added to the accounts file.
To print lines 6-10:
sed -n '6,10p' file If the file is huge, and the end line number is small compared to the number of lines, you can make it more efficient by:
sed -n '10q;6,10p' file From testing a file with a fairly large number of lines:
$ wc -l test.txt 368048 test.txt $ du -k test.txt 24640 test.txt $ time sed -n '10q;6,10p' test.txt >/dev/null real 0m0.005s user 0m0.001s sys 0m0.003s $ time sed -n '6,10p' test.txt >/dev/null real 0m0.123s user 0m0.092s sys 0m0.030s If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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