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# 1  
Old 07-03-2017
sed
Does anyone know how I can do the following?

I have multiple blocks of text, I enclose an example of 2.
1 has the word DRAFT in the 2nd line and will always be in the 2nd line.
How can I remove all blocks of text that have DRAFT in them?

sedl


Code:
SAMPLE SOME TEXT AND MORE TEXT 
DRAFT
SAMPLE TEXT
SAMPLE TEXT
BLOCK ONE
END

SAMPLE SOME TEXT AND MORE TEXT 
SAMPLE TEXT
SAMPLE TEXT
BLOCK TWO
END

Many thanks...


Moderator's Comments:
Mod Comment Please use CODE tags as required by forum rules!
Last edited by RudiC; 07-03-2017 at 07:37 PM.. Reason: Added CODE tags.
# 2  
Old 07-03-2017
Hi,
you can try this if you use gnu sed:
Code:
sed -n ':w;/END/bc;$bc;N;bw;:c;/^[^\n]*\nDRAFT/!p' file

Regards.
# 3  
Old 07-03-2017
Try also
Code:
awk '!($2 ~ /DRAFT/)' FS="\n" RS= file

# 4  
Old 07-04-2017
Quote:
Originally Posted by disedorgue
Hi,
you can try this if you use gnu sed:
Code:
sed -n ':w;/END/bc;$bc;N;bw;:c;/^[^\n]*\nDRAFT/!p' file

Regards.
A Unix sed needs the labels at the end of a line (semicolon not allowed there)
Code:
sed -e ':w' -e '/END/bc' -e '$bc' -e 'N;bw' -e ':c' -e '/\nDRAFT/d' file

This version allows DRAFT on any line within the block.
# 5  
Old 07-04-2017
Quote:
Originally Posted by RudiC
Try also
Code:
awk '!($2 ~ /DRAFT/)' FS="\n" RS= file

I think you would also need to set ORS otherwise it will remove all the empty lines:
Code:
awk '$2!~/DRAFT/' RS= FS='\n' ORS='\n\n' file

This User Gave Thanks to Scrutinizer For This Post:
RudiC
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