How to Set the Grep After Context to Be "Until the Next Blank Line"

Grep for next blank line after line number

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

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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

Obtain all the lines before and after a pattern match until a there is blank line

Use awk's paragraph mode (one or more empty lines act as record separator)

$ awk -v RS= '/ACCCC/' ip.txt 
ABASLDKJ
ASDASKKK
ASDASDAS
ACCCC
ASDASDAS
ASDASDAS

• -v command line option helps to set value to a variable
• RS is input record separator, whose default value is newline character

From awk manual

By a special dispensation, an empty string as the value of RS
indicates that records are separated by one or more blank lines. When
RS is set to the empty string, each record always ends at the first
blank line encountered. The next record doesn’t start until the first
nonblank line that follows. No matter how many blank lines appear in a
row, they all act as one record separator. (Blank lines must be
completely empty; lines that contain only whitespace do not count.)

How to print only the line which contains a blank lines before and after

With your shown samples, could you please try following.

awk -v RS="" 'NF==1' Input_file
tuv0657
tuv2330
tuv1049
tuv2101

Match empty lines in a file with 'grep'

The regular expression to match the end of the line is $, not \n (since grep works a line at a time, it ignores the newlines between lines).

grep -c '^$' myfile.txt

grep: show lines surrounding each match

For BSD or GNU grep you can use -B num to set how many lines before the match and -A num for the number of lines after the match.

grep -B 3 -A 2 foo README.txt

If you want the same number of lines before and after you can use -C num.

grep -C 3 foo README.txt

This will show 3 lines before and 3 lines after.

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