Question

I have a large file which contains tables of different tab separated data. The different tables are separated by a blank line.

I have the line number of the start of a particular table and I need to retrieve the whole table.

How can I use grep (or something similar) to get the line number of the next blank line after a specific line number?

Answer 1

Use sed for this, this should do the trick:

sed -n '1,/^\s*$/p' file

Just replace the first number before the comma, in this case 1 with the line number, demo to print each table from a given line number:

$ cat file
one
two
three

five
six
seven

nine
ten
eleven

$ sed -n '1,/^\s*$/p' file
one
two
three

$ sed -n '5,/^\s*$/p' file
five
six
seven

$ sed -n '9,/^\s*$/p' file
nine
ten
eleven

Using the -n option to turn of default printing of every line and the p flag sed prints from the line number to the first line that matches the regexp where:

^     # Matches the start of the line
\s*   # Matches zero or more whitespace characters
$     # Matches the end of the line

Using the format sed -n 'A,Bp' where A and B can be either lines numbers or regular expression you can print subsections of files easily.

To print just the line number of the next blank line with sed do:

$ sed -n '1,/^\s*$/{=}' file | tail -1
4

$ sed -n '5,/^\s*$/{=}' file | tail -1
8

$ sed -n '9,/^\s*$/{=}' file | tail -1
12

Or just printing where all the blanks lines are

$ sed -n '/^\s*$/{=}' file
4
8
12

---

Getting the next blank line number with awk doesn’t require using tail:

$ awk 'NR>=1 && /^\s*$/{print NR;exit}' file
4

$ awk 'NR>=5 && /^\s*$/{print NR;exit}' file
8

$ awk 'NR>=9 && /^\s*$/{print NR;exit}' file
12

$ awk '/^\s*$/{print NR}' file
4
8
12

If it makes it clearer for you, you can pass a variable in with awk using -v

$ awk -v start=1 'NR>=start && /^\s*$/{print NR;exit}' file
4

$ awk -v start=5 'NR>=start && /^\s*$/{print NR;exit}' file
8

$ awk -v start=9 'NR>=start && /^\s*$/{print NR;exit}' file
12