awk : if >=4 lines in a row begin with + or - don't print record
If I understood your question, the input file have records as paragraphs, so you will need to separate them with blank lines. I assumed it for next script:
Content of script.awk:
BEGIN { ## Separate records by one or more blank lines. RS = "" ## Each line will be one field. Both for input and output. FS = OFS = "\n"}## For every paragraph...{ ## Flag to check if I will print the paragraph to output. ## If 1, print. ## If 0, don't print. output = 1 ## Count how many consecutive rows have '+' or '-' as first ## character. j = 0 ## Traverse all rows. for ( i = 1; i <= NF; i++ ) { if ( substr( $i, 1, 1 ) ~ /+|-/ ) { ++j; } else { j = 0 } if ( j >= 4 ) { output = 0 break } } if ( output == 1 ) { print $0 "\n" }}Assuming following test input file as infile:
+Host is up. +Not shown: 95 closed ports, 3 filtered ports+PORT STATE SERVICE VERSION+Host is up. +Not shown: 95 closed ports, 3 filtered ports+PORT STATE SERVICE VERSION+23/tcp open telnet+9100/tcp open jetdirect-Host is up. -Not shown: 99 closed ports-PORT STATE SERVICE VERSION-5900/tcp open vnc -Not shown: 76 closed ports, 18 filtered ports+Not shown: 93 closed portsPORT STATE SERVICE VERSION+514/tcp open shellRun the script like:
awk -f script.awk infileWith following output (first record because it doesn't reach to four consecutive rows, and second record because it has a different line between them):
+Host is up.+Not shown: 95 closed ports, 3 filtered ports+PORT STATE SERVICE VERSION-Not shown: 76 closed ports, 18 filtered ports+Not shown: 93 closed portsPORT STATE SERVICE VERSION+514/tcp open shell