How to check if 3 files have same size in directory


 
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# 1  
Old 02-15-2005
How to check if 3 files have same size in directory

I need to determine if any three files have the same file size in a specified directly?

I have got as far as listing the file sizes but where to go from here?

ls -al |sort -n -r +4 | awq '{print $5}'

Thanks in anticipation
# 2  
Old 02-15-2005
Code:
ls -ltr x1 x2 x3 |
awk 'BEGIN { sum=0 ; value=0}
{ sum+=$5 ; value=$5 } END { print sum ; print sum/3 ; if ( sum/3 == value )
  {
     print "Equal" ;
  }else
  {
     print "Not Equal"
  }
}'

# 3  
Old 02-16-2005
How to check if any three files are the same size in a directory

If I am not mistaken this assumes you know the names of the three file to compare.

What I need is a script that assumes you do not know the names "x1" "x2" "x3" at the start.

Say you have a directory of 1000 files - the question is do any three (or more) of these 1000 files have the same size.

or another way to satisfy my needs would be a script:

Do the last three files created have the same size?

Thanks.
# 4  
Old 02-16-2005
Quote:

Say you have a directory of 1000 files - the question is do any three (or more) of these 1000 files have the same size.
For the above ... pass aruguments to the script $1,$2,$3 ... and so on
You wil know how many number of arguments passed.

Then "x1 x2 x3" in the script will be replaced by $*

ls -ltr $*

Devide sum by $# ( no of files you are passing to the script)
the remaining script will be same.



Quote:
Do the last three files created have the same size?
It is very easy to find the last three files created
Code:
ls -lt | awk '{ (if NR > 1 && NR < 5 ) print $0 }'


Hope it helps ....
# 5  
Old 02-16-2005
My solution - last three created files same size?

Thanks for this - final line is:

ls -l | tail -3 | awk 'BEGIN {sum=0 ; value=0} { sum+=$5 ; value=$5 } END { if ( sum/3 == value ) { print "equal" ; } else {print "not equal" } }'

on the assumption ls -l returns the files in created time sequence by default
# 6  
Old 02-16-2005
Is it not ls -lat instead of ls -al ?
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