tadakamalla

Iam working on Free BSD 5.3 version. U all know that the output of 'df' command looks as follows:

Filesystem 1K-blocks Used Avail Capacity Mounted on
/dev/ad0s1a 253678 59116 174268 25% /
devfs 1 1 0 100% /dev
/dev/ad0s1e 253678 24280 209104 10% /tmp
/dev/ad0s1f 8913022 2615376 5584606 32% /usr
/dev/ad0s1d 253678 30246 203138 13% /var
devfs 1 1 0 100% /var/named/dev


Now i want to check which partition if greater than 90% full without scanning the first row of record. How to do this using awk command? PLease anyone of u help me with this?

muha

depending on what you are trying to do:
df |sed '1d'|awk '{if ($5 > 90) print $5,$6}'|sort -r
99% /harddisk/C
96% /harddisk/E

df |sed '1d'|awk '{if ($5 > 90) print $6}'|sort -r
/harddisk/C
/harddisk/E

or unsorted:
df |sed '1d'|awk '{if ($5 > 90) print $5,$6}'
96% /harddisk/E
99% /harddisk/C

/edit: df |sed '1d'| awk '$5 > 90 {print $5,$6}'
also works ..

bryansmith

Does this have anything to do with your homework sounding question?

Bryan

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tadakamalla

Its not working Muha, i have tried urs but is not working.

And actually i have to compare the 5th field with greater than 90% from the output o fthe df command. So give me a clue to accept this % symbol.