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I am trying to get data after the n:th line number.

I have a file with 1500 lines, but I want to print the data after 750 lines.

I tried head and tail, but couldn't get exactly what I wanted.

Kusalananda
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Praveen
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  • "print"? To the console? On a piece of paper? – DeeJayh Sep 20 '21 at 18:31
  • @DeeJayh It is customary to use the verb "print" to mean "to output". Would they want to actually get a hard copy of the output, then it would be trivial to pipe the result to lp or equivalent print spooler command. This does not change the what question is asking for. – Kusalananda Sep 20 '21 at 18:34
  • I thought tags will be checked, my bad. Trying in bash on linux server – Praveen Sep 20 '21 at 18:35

2 Answers2

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You can use tail

tail -n +751 file

from man tail:

use -n +NUM to output starting with line NUM

Alternative using sed:

sed '1,750d' file

(delete everything from line 1 to 750)

pLumo
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The awk command can be used with operators to specify the number record. In this case greater than (after) record 750.

awk 'NR>750' input_file_name

Detailed explanation

The awk command, or its distribution-specific counterparts like mawk in ubuntu, are usually available even in the most lean, base distrubitions. An awk program is a sequence of patterns and corresponding actions. The awk program 'NR>750' simply returns all records after line number 750.

Sourced from: https://stackoverflow.com/a/25678925/5387389

DeeJayh
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    I think you need NR>750 as OP wants to print after line 750. Maybe that is why the downvote. However, that should be a comment rather than a downvote. – pLumo Sep 20 '21 at 18:47
  • @pLumo that could be right. I'll update the answer. – DeeJayh Sep 20 '21 at 18:52