Shell Programming and Scripting

Hi

I have read your article on grep it is very useful.

I have a problem using grep , can you please help me to sort out.

I am listing the contents of a file below :

##########1003134 05/04/1970F48371008 0736778 6123785090 0E10/13/1997

04/22/1998XSSUB 77 37 .00 -26.52

##########1004826 06/07/1949M58788224 0780015 6120000000 0E03/26/1998

XCIA 77 33.20 -251.65 W 12

##########1742034 04/05/1919F47094272 0627422 6126318148 0E04/28/1997

XXEA 77 .00 -131.92 P ##########1753219 08/13/1922F47185401 0022198 6126233676 0E06/04/1997

XCIA 77 23201DMN .00 -18.33
##########1759194 05/20/1957F47640131 0741645 6127224265 0E06/24/1997

XXUA 77 120.80 .00 W ##########1760725 03/22/1914F46013107 1069089 6128275931 0E06/29/1997

XXEA 77 147.00 -86.97 I 090901 ##########1761672 12/25/192M129341331 1055160 6128706104 0E07/02/1997

XXEA 77 171.18 -359.58 X

I need to remove some junk lines which appear before the first line shown above, I have not listed the junk lines here. I need to remove the

Junk lines and send the valid lines ie the lines which start with 10 hash (#) to an output file, so I have used the following command.


$ grep -E “\ # \ # \ # \ # \ # \ # \ # \ # \ # \ #” testfile.txt > mainfile.txt

My intention here is to take out all the lines which start with 10 #'s in to the other file, and as you see that each row has three lines and when

I use this command it is just picking the line which has the hash and skipping the other 2 lines.

Is there any solution for this problem using grep or do I need to use any other method.

Thanks

Vittal Raj

Just out of curiosity, what article on grep are you referring to..?

Is this file really how your file looks? The entire file looks unreadable.. To me, it doesn't look like every line with hashes in it is followed by two lines.. or is this how you meant for it to look?

Code:

##########1003134 05/04/1970F48371008 0736778 6123785090 0E10/13/1997

04/22/1998XSSUB 77 37 .00 -26.52

##########1004826 06/07/1949M58788224 0780015 6120000000 0E03/26/1998

XCIA 77 33.20 -251.65 W 12

##########1742034 04/05/1919F47094272 0627422 6126318148 0E04/28/1997

XXEA 77 .00 -131.92 P

##########1753219 08/13/1922F47185401 0022198 6126233676 0E06/04/1997

XCIA 77 23201DMN .00 -18.33

##########1759194 05/20/1957F47640131 0741645 6127224265 0E06/24/1997

XXUA 77 120.80 .00 W

##########1760725 03/22/1914F46013107 1069089 6128275931 0E06/29/1997

XXEA 77 147.00 -86.97 I 090901

##########1761672 12/25/192M129341331 1055160 6128706104 0E07/02/1997

XXEA 77 171.18 -359.58 X

grep -E "#{10}" will move all the lines with hashes in them to another file, but I'm not sure what else you want to pull out..

I think that he wants the line that starts with ########## and whatever two lines happen to follow it. Try this:

sed -n '/^##########/ { N ; N ; p ; } '

Yes Perderabo got it right.

I am new to shell scripting.

I have a question can i use the command you have sent

by giving the inputfile and redirecting it to a outputfile like

sed -n '/^##########/ { N ; N ; p ; } ' inputfile.txt > outfile.txt

I am sorry if i am irritating you with questions.

Thanks for the time!

Vittal Raj

Yes, that should work fine.

Just adding to the quetion above, What if a file consist of records, where in each records consist of multiple line, and each new records starts with a #.

If the number of lines followed by # is fixed, then it can be easily solved by the answer given above. But what if the number of lines in a records is not fixed? How to extract then?

Regards,
Yeheya

There is not enough information here to answer the question. If a line like "#start" is followed 30 lines and only some of them are continuations of the record, the question is how do you tell? If there is a way to tell where the record ends, that way can be programmed. And we also need to know which records to select and which to ignore.

If you post a concrete example, then we can give a concrete solution.