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Need Shell Script for Array

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# 8  
Old 02-23-2015
Hello Hemanth,

Following are the answers for your queries.
1: yes sort -r -k 3,3 means to sort the text in reverse order by taking the 3rd column as primary key.
2: yes, as I have mentioned in my very first post this code will work only on 3 fields seperated with ,.


Hope this helps, let us know if you have any queries on same. If you have exact data in a different format please do let us know always from the very first post itself, it helps us too to help/guide.


Thanks,
R. Singh
# 9  
Old 02-23-2015
With a simple input structure as in your sample, no matter how many lines (records) the file will contain, this might work:
Code:
awk '{print "record", $1, "sum is", $2; print "record", $1, "count is", $3; }' FS="," file
record  345 sum is 12
record  345 count is 10
record  400 sum is 11
record  400 count is 8
record  328 sum is 1
record  328 count is 3

For the order that you require, pipe it through sort as RavinderSingh13 proposed.
# 10  
Old 02-23-2015
If all of the following are true:
  • you want the output order to be the same as the input order for both "sum" lines and "count" lines,
  • you don't really have leading spaces in your input file, and
  • you don't really want spaces at the start of each output line after the first output line,
you could try something like:
Code:
awk -F, '
{	printf("record %s sum is %s\n", $1, $2)
	r[NR] = $1
	c[NR] = $3
}
END {	for(i = 1; i <= NR; i++) printf("record %s count is %s\n", r[i], c[i])
}' file

which produces the output:
Code:
record 345 sum is 12
record 400 sum is 11
record 328 sum is 1
record 345 count is 10
record 400 count is 8
record 328 count is 3

Note that the input line:
Code:
400,11,8

in your sample input produces the output shown in red above, not the output:
Code:
record 400 sum is 10

that you said you wanted. If you really want the line above, you need to clearly explain what transformations are supposed to be applied to the data in the 2nd input field in your input records to produce the output you say you want.

If you want to try this script on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk or nawk.
# 11  
Old 02-23-2015
hi ,
thanks for the reply..i just gave a example for the input file,

Can you please tell me like how to print the sum digit in currecy format ..?

like
Code:
record 345 sum is 12,345

Consider like in the input file the 345 record sum is 12345
i know that awk command do...
Code:
awk -F "," '{printf(fmt,value)}' FS=,OFS=, fmt="%'.2f"

can you please tell me like how to do that in the code you mentioned i didnt get that %s concept in the code

thanks in advance
sai
# 12  
Old 02-23-2015
You'll have to divide the number by 100 and replace the "%s" with your desired format "%.2f". Note that not all versions will recognize the " ' " in the format string.
# 13  
Old 02-23-2015
hi ,
in the below code
Code:
awk -F, '
{ printf("record %s sum is %s\n", $1, $2)
r[NR] = $1
c[NR] = $3
}
END { for(i = 1; i <= NR; i++) printf("record %s count is %s\n", r[i], c[i])
}' file

how can i represent to get the $2 value as currency format as in awk command its not accepting "%'.2f" as it contains ' symbol... can u help me on this like wat symbol i should add in the place of %s to get the desired format like 123,456 , 1,234 ..... Smilie
# 14  
Old 02-23-2015
Why don't you do it stepwise - like print the number divided by 100, print it with the "%.2f" format, and so on - and present us with the intermediate results so we see what goes wrong when.

As I said, the " ' " is not accepted on all systems/tools. Looks like yours is one of those. You'll now have to work on the resulting string if you insist on the thousands' separator.
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