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help one liner...


 
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# 1  
Old 06-09-2011
help one liner...

Hi I have a log data that shows chunks of data like this:

thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example testthisis example test, this is example test, this is example test
12
6 thisis example test, this is example test, this is example test
6 thisis example test, this is example test, this is example test

thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
thisis example test, this is example test, this is example test
2500
1000 thisis example test, this is example test, this is example test
1000 thisis example test, this is example test, this is example test
200 thisis example test, this is example test, this is example test
200 thisis example test, this is example test, this is example test
100 thisis example test, this is example test, this is example test


I am interested in printing the chunk of data that contains a number (lonely number) larger than 2000.

The chunk always end on a new line. but the size of the chunk is variable and I need to focus on the lonely number>2000 but print the whole chunk.

Please only use bash commands, and awk,sed or grep. I don't want python or perl. Thanks
# 2  
Old 06-09-2011
Code:
awk '/^[0-9]+$/ { if ($0 > 2000) print }'

# 3  
Old 06-09-2011
This will print all the lines where there is a lonley number which is greater than 2000

Code:
 
awk 'chunk=$0{gsub(/[^0-9]/,"");{if($0> 2000){print chunk}}}'  filename


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