Grep out an entire content up to a blank line (as a delimiter)

victordcweb · 10-18-2016, 12:27 PM

I'm trying to grep out an entire content up to a blank line. Any help will be appreciated. I'll prefer a grep command but I'm okay with other solution

Have: File

TableA has the following
GET TABLE TABLEA
(value(1)
)
/

TABLEB has the following
GET TABLE TABLEB
(value(1)
,value(2)
)
/

WANT

An output to contain the following

TABLEA has the following
GET TABLE TABLEA
(value(1)
)
/

schneidz · 10-18-2016, 12:37 PM

i would process the file twice. the first time to grep the line number of the first blank line; then 2nd use head to print up to that line.

grail · 10-18-2016, 12:55 PM

awk would work just fine if you set the RS delimiter to your blank line

MadeInGermany · 10-18-2016, 02:35 PM

With awk:

Code:

awk 'NF==0 {pr=0} /TABLEB/ {pr=1} pr' file

or passed as a parameter

Code:

awk 'NF==0 {pr=0} $0~search {pr=1} pr' search="TABLEB" file

Turbocapitalist · 10-18-2016, 03:06 PM

And just to throw in a "sed" variant, you could try

Code:

sed -n '/^TABLEA/,/^$/{/^$/!p}' < infile.txt

Just remember that it is case sensitive.

sed -n '/A/,/B/{/C/!p}'

The -n means by default don't print anything
The /A/,/B/ { } means from pattern A to pattern B, inclusive, do what's in the braces.
The /C/!p means except when pattern C occurs, print the current pattern space (line)

keefaz · 10-18-2016, 05:29 PM

With Grep and Perl regexp extension:

Code:

grep -Pzo '(?s)TableA.*?\n$' file.txt

Edit: not sure if you want to grep 'TableA' or 'TABLEA'
Maybe add i Perl modifier

Code:

grep -Pzo '(?si)TableA.*?\n$' file.txt

AwesomeMachine · 10-18-2016, 06:31 PM

The last character (nonprinting) of a line before a blank line is newline "\n". You should use awk to print everything up to the first newline character.

grail · 10-19-2016, 04:58 AM

Code:

awk '/TABLEA/' RS="" file