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awk help

 
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# 1  
Old 02-16-2009
awk help
Hi,
I just started using linux, and I am trying to use awk to search for a string in a document. If it exists I would like to print the next 33 lines(there will be mulitple matches in these files). Ive tried variations of this form, but dont know how to print by line.


awk '{if ($4=1970) print ???? }' oldfile>newfile.dat

Thanks for any tips
# 2  
Old 02-16-2009
Try:

Code:
awk 'c&&c--;/pattern/{c=33}' infile

If you want to include the matching line:

Code:
awk '/pattern/{c=33}c&&c--' infile

Add one to c if you want the matching line and the following 33 lines (total 34 lines).
# 3  
Old 02-16-2009
Opps, I forgot to mention the first field for this line needs to say "start" as well can this be added in to that search pattern?
Let me try and clarify...
I am looking for these two things in the line and if they match I need all the lines after that until the next "start" This is much harder than I first thought.
start xxx xxx 1970

sorry about the initial confusion
# 4  
Old 02-16-2009
It would help if you posted a sample data file using BB Codes and a desired output.

Assuming there's no matching 'stop' for 'start':
Code:
nawk '/^start.*1970/ {c++;print "--";next} c'

# 5  
Old 02-16-2009
This might be a dumb question but I dont think I have the nawk command. Can this be run using just awk.
I am new to our system and it says nawk command not found
# 6  
Old 02-16-2009
This is the format of the file. The output can be the same. I just need to extract the entire line with "start", and a value in the fourth column, and all of the following info until the next "start" for the entire file. Im sorry this might be confusing, I am very new to this.

Code:
start     -71.50000     40.37000    1981       1       7       8       1
      0.00000      7.43000   9999.00000
     25.00000      7.50000   9999.00000
     45.00000      7.51000   9999.00000
     60.00000      7.68000   9999.00000
     64.00000      8.72000   9999.00000
     66.00000     10.13000   9999.00000
     69.00000     10.82000   9999.00000
     81.00000     10.83000   9999.00000
start     -71.52700     40.15500    1981       1       7       7       2
      0.00000      9.51000   9999.00000
      4.00000      9.39000   9999.00000
     35.00000      9.39000   9999.00000
     64.00000      9.46000   9999.00000
     68.00000     10.69000   9999.00000
     70.00000     10.94000   9999.00000
     85.00000     10.98000   9999.00000

I hope this comes through alright
Last edited by vgersh99; 02-16-2009 at 02:21 PM.. Reason: BB code tags
# 7  
Old 02-16-2009
Quote:
Originally Posted by d_kowalske
This is the format of the file. The output can be the same. I just need to extract the entire line with "start", and a value in the fourth column, and all of the following info until the next "start" for the entire file. Im sorry this might be confusing, I am very new to this.

start -71.50000 40.37000 1981 1 7 8 1
0.00000 7.43000 9999.00000
25.00000 7.50000 9999.00000
45.00000 7.51000 9999.00000
60.00000 7.68000 9999.00000
64.00000 8.72000 9999.00000
66.00000 10.13000 9999.00000
69.00000 10.82000 9999.00000
81.00000 10.83000 9999.00000
start -71.52700 40.15500 1981 1 7 7 2
0.00000 9.51000 9999.00000
4.00000 9.39000 9999.00000
35.00000 9.39000 9999.00000
64.00000 9.46000 9999.00000
68.00000 10.69000 9999.00000
70.00000 10.94000 9999.00000
85.00000 10.98000 9999.00000

I hope this comes through alright
Try this:

Code:
awk '
/start/ {p=1;print $4;next}
/start/ && p {exit}
p' file

Regards
 
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