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Replicating many rows in unix

 
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# 1  
Old 12-08-2011
Replicating many rows in unix
Hi

If my input is

Code:
abcdefgh
ijklmnop

then

the output should be:


Code:
abcdefgh
abcdefgh
abcdefgh
abcdefgh
ijklmnop
ijklmnop
ijklmnop
ijklmnop

i.e.One row in the input becomes 4 records in the output.How we can achieve this in Unix?
# 2  
Old 12-08-2011
Many different ways, here's one:

Code:
awk ' { for( i = 0; i < 3; i++ ) print; }' input-file-name

These 2 Users Gave Thanks to agama For This Post:
pandeesh Siva Sankar
# 3  
Old 12-08-2011
Fine.

If you want to add some characters with replication:
in my case,if i need the output as:

Code:
abcdefgh12
abcdefgh15
abcdefgh17
abcdefgh19
abcdefgh21
ijklmnop12
ijklmnop15
ijklmnop17
ijklmnop19
ijklmnop21

Those 12,15,17,19,21wil be always constant.

How we can achieve this?

Thanks
# 4  
Old 12-08-2011
Easiest way if they are all constant:

Code:
awk 'BEGIN { split( "12 15 17 19", add, " " ); } { for( i = 1; i < 5; i++ ) print $0 add[i]; }'

This User Gave Thanks to agama For This Post:
pandeesh
# 5  
Old 12-08-2011
Thanks Agama!

So i can use:

Code:
awk 'BEGIN { split( "12 15 17 19 21", add, " " ); } { for( i = 1; i < 5; i++ ) print $0 add[i]; }'

# 6  
Old 12-08-2011
Yep. What ever you pass in the string to split() will be added to the end of the respective lines of output, in the order. The numbers can be text, as long as you separate them with spaces. Easy to change if you need spaces, but I think that will work for you as is.
# 7  
Old 12-08-2011
Actually as per my requirement, i need to add those values in a comma delimited record as last field.
I believe the below will work in that case:
Code:
awk 'BEGIN { split( ",12 ,15 ,17 ,19 ,21", add, " " ); } { for( i = 1; i < 5; i++ ) print $0 add[i]; }'

Thanks
 
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