UNIX for Dummies Questions & Answers

Hi everyone,

I am writing a script. As a part of this script, I wanted to check the space of few mounts, If the space usage percentage of the mount crosses over a certain limit then i wanted to display a warning message. The following is the command

df -k | awk ' { if (($6 == "/export/temp") || ($6 == "/export/temp1") || ($6 == "/export/temp2")) {svalue=substr($5,1,(index($5,"%"))-1);if(svalue >= 48) { print "\nWarning: "$6 " space is " $5 " running out of space"}}}'

when the percentage of space is equal to 100% it is not printing the warning message. If i am not wrong, i think the return value of "substr($5,1,(index($5,"%"))-1" is a string which will be stored in svalue string variable. Then i am comparing the string value of svalue with a interger value. This may be the problem I think so.

can anyone tell me how to correct it? how to convert the string value to integer value and then compare it? or how to do it in a correct way?

Please help me on this.
Thank in advance.

awk automatically converts between string and number representation, and complains if it cannot do that. I believe you could even skip the whole substr thing, and trust awk to parse the number part before the percent sign.

It would perhaps be more idiomatic to use a regular expression for $6, and make it the condition for the rest of the script.

Code:

df -k | awk '$6 ~ /^\/export\/temp/ && $5+0 >= 48 {
    print "Warning: "$6 " space is " $5 " running out of space"}'

I'll leave it to you to ponder on whether "space is running out of space" is a useful thing to say ...

(If you have other /export/temp* partitions than the ones you listed, you'll have to tighten up that regular expression a bit. /^\/export\/temp[12]?$/ would match exactly what you had.

Hi ,

I tried as you have said but it is not printing anything,

df -k | awk '$6 ~ /^\/export\/home1/ && $5+0 >= 48 {print "Warning: "$6 " space is " $5 " running out of space"}'

so just to check whether awk parse the number part before the percent sign, i have executed the following command

df -k | awk ' { if (($6 == "/export/home1") || ($6 == "/export/home2")) {svalue=$5+0;print "svalue=" svalue;print "per of space=" $5;}}'

i am getting the output,
svalue=0
per of space =65%
svalue=0
per of space=49%

Any advice please?

Works for me with both mawk and classic awk. Which platform are you on?

If the match you did before didn't work correctly, and this doesn't work, you could try a simple sub(/%/, "", $5) to get rid of the percent sign.

Hi,

Thanks for your quick response.

I am using SunOS 5.8 version. I have tried as you have suggested like

df -k | awk '($6 ~ /^\/export\/home1/) && (sub(/%/, "", $5) >= 48) {
print "Warning: "$6 " space is " $5 " running out of space"}'
then i am getting the syntax error,
awk: syntax error near line 1
awk: bailing out near line

I think it may be a problem since i am trying to change to $5 instead of string variable value,

so i have changed the code to
df -k | awk ' { if ($6 == "/export/home1") {svalue=$5;sub(/%/, "", svalue);print svalue}}'
then also i am getting the same error,
awk: syntax error near line 1
awk: illegal statement near line 1

I am not sure what i am doing wrong. Please advice me on this.

Then i have modified the code the other way,

df -k | awk ''(($6 == "/export/home1") || ($6 == "/export/home2")) {if((substr($5,1,(index($5,"%")-1))+0) >= 5) { print "\nWarning: "$6 " space is " $5 " low Disk space"}}'

The above code is working for me even when the $5 i.e usage percentage of space is 100% also.

I am not sure what was wrong with the earlier code. Please advice.