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How to takes the missing files?

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# 1  
How to takes the missing files?

Hi all ,
am using unix aix
I have a files in one directory..
my files as in format qqss0607.ddd.. (06 is date 07 is month)

how to check the missing dates ....
can anyone tell me...
# 2  
seq 0 365|while read line 
    d=$(date -d "Jan 1 +$line days " +%d%m)
    if [[ ! -f  $filename ]]; then
         echo "$filename is not exist"

This User Gave Thanks to rdcwayx For This Post:
# 3  
The date -d trick only works with GNU date, which AIX likely doesn't have. You may be able to do something similar with perl.
# 4  
Solaris has no GNU date also, but install a GNU date in unix system is not big deal.
# 5  
Give this a go (dont think seq is available in AIX either):

set -A days 31 $(cal 02 $(date +%Y) | awk 'NF{E=$NF}END{print E}') 31 30 31 30 31 31 30 31 30 31
for mth_days in ${days[@]}
   let mth=mth+1
   while [ $day -le $mth_days ]
       fname=$(printf "qqss%02d%02d.ddd" $day $mth)
       [ -f $fname ] || echo $fname
       let day=day+1

This User Gave Thanks to Chubler_XL For This Post:
# 6  
thanks chubler...
i got the results per years fully (its not checking the files displaying full yrs with filename )

I need the outputs only for missing files per month...
for ex.. i have files like

i need output as
missing files qq0201.d

---------- Post updated at 12:48 AM ---------- Previous update was at 12:47 AM ----------

Thanks rdcwayx..
am using unix ksh
seq wont works ...
# 7  


Try this one,
perl -alne 'my $path="/home/path/";my @day=(31 28 31 30 31 30 31 31 30 31 30 31);for(my $i=1;$i<=12;$i  ){for(my $j=1;$j<=$day[$i-1];$j  ){my $d=sprintf("%02d%02d",$j,$i);my $file=join('',"$path"'qqss',"$d");if(! -f "$file"){print "FILE NOT FOUND;$file";}}}'


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