Else if in awk


 
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# 1  
Old 10-28-2015
Else if in awk

Hi All,
my input file is comma delimited.
Code:
1,Y,1111,9999
2,Y,1,9999
2,N,1,9999
2,Y,9999,9999

o/p should be
Code:
1,Y,1112,9999
2,Y,0002,9999
2,Y,0000,9999

scenario is
1.if $2 is "N" then exit.
2.if $3 is not equal to $4 then increment $3 by 1 padding to four decimal places to the left.
3.if $3 is equal to $4 then reset it zero padding with 4 zero.

my attempt for above 2 scenario is:-
awk -F, '{if ($2=='N') {exit 1} else if( $3!=$4) { $3+=1} {$3= sprintf("%04d",$3) }{print $0 } }' test
and it working

but after including scenario 3 it is giving syntatical error:-
awk -F, '{if ($2=="N") {exit} else if( $3!=$4) { $3+=1} {$3= sprintf("%04d",$3) } else if ($3=="9999") {$3="0000"} {print $0 }}' test
Kindly help me on this.
Thanks,

Last edited by looney; 10-28-2015 at 09:28 AM..
# 2  
Old 10-28-2015
Hello looney,

Could you please try following and let me know if this helps you.
Code:
awk -F, '($2=="N"){next} ($3==$4){$3=0} ($3!=$4){$3=$3+1} {printf "%d,%s,%04d,%d\n",$1,$2,$3,$4}'  Input_file

Output will be as follows.
Code:
1,Y,1112,9999
2,Y,0002,9999
2,Y,0001,9999

Off course it is according to input shown by you to us(Like input file have 4 columns only and it is , delimited), if you have more conditions please do let me know on same.


Thanks,
R. Singh

Last edited by RavinderSingh13; 10-28-2015 at 09:41 AM..
# 3  
Old 10-28-2015
scenario
1. will be met if you double quote the "N" (instead of single quoting). Pls note that line 4 will not be read/printed if script exits on line 3, and your output wouldn't match this.
2. is not met in your second snippet as the condition $3 == "9999" must also be met.
3. doesn't mention you want $3 to be four digits zero padded.
# 4  
Old 10-28-2015
Hi Ravinder
for 3rd record , 3rd field should be 0000 as in below
2,Y,0000,9999. The logic is whenever 3rd field reaches to 9999 it should be again reset to 0000
can we do it by if else condition mentioning explicitly.?
Thanks,
# 5  
Old 10-28-2015
Try
Code:
awk -F, '
                {$3=sprintf ("%04d", ($3 == $4)?0:$3+1)
                }
$2!="N"
' OFS=, file

# 6  
Old 10-28-2015
Hello looney,

Could you please try following and let me know if this helps. Let's say we have following Input_file.
Code:
cat Input_file
1,Y,1111,9999
2,Y,1,9999
2,N,1,9999
2,Y,9999,9999
2,Y,9999,0002
3,N,2343,0002
4,Y,1203,1203

Then following is the code for same.
Code:
awk -F, '($2=="N"){next} ($3==$4 || $3==9999){$3=0} ($3!=$4){$3=$3+1} {printf "%d,%s,%04d,%04d\n",$1,$2,$3,$4}'   Input_file

Output will be as follows.
Code:
1,Y,1112,9999
2,Y,0002,9999
2,Y,0001,9999
2,Y,0001,0002
4,Y,0001,1203

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
# 7  
Old 10-28-2015
For the wrap around at 9999, try
Code:
awk -F, '
                {$3=sprintf ("%04d", ($3 == $4)?0:($3+1)%1E4)
                }
$2!="N"
' OFS=, file

This User Gave Thanks to RudiC For This Post:
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