awk to print line greater than zero in column 1

Question

I am trying to print the name of directories whose free space is greater than zero.

  #Listed all the directory and their consumed space.
    du -sm storage*
    365     storage1
    670     storage2
    1426    storage3

I have threshold value of 1000M , so I am trying to print free space in these directories relative to threshold value provided.

du -sm storage* | awk -v threshold="1000" '$1>0{print $1=threshold-$1,$2}'
635 storage1
330 storage2
-426 storage3

So , I want to print those directories whose free size is positive integer. Something like :

635 storage1
330 storage2

Any correction ?

Answer 1

You can write it like this,

awk -v threshold="1000" '{$1=threshold-$1} $1 > 0'

Example

awk -v threshold="1000" '{$1=threshold-$1} $1 > 0' input
635 storage1
330 storage2

What it does?

• $1=threshold-$1 Sets the first column relative to the threshold.

• $1 > 0 Checks if the derived first column is greater than zero. If this expression evaluates true, it prints the entire input line.

Answer 2

I think this got overly complicated. If you just want to check that the size is positive and lower than a given threshold, say so:

awk -v threshold=1000 '0 < $1 && $1 < threshold'

Test

$ cat file
635 storage1
330 storage2
-426 storage3

$ awk -v thr=1000 '0 < $1 && $1 < thr' file
635 storage1
330 storage2

Answer 3

Just check $1 > 0 in awk

du -sm storage*|awk '{if ( $1 > 0 ) print }'

or

du -sm storage*|awk  '$1>0'