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Old 12-17-2016, 05:49 AM   #1
limdel
Member
 
Registered: May 2009
Location: Ho Chi Minh City, Vietnam
Distribution: CentOS - FEDORA
Posts: 59

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script to test IOPS on multiple mounted disk


Hi,
My goal is to have a single script that would output something like this:
Code:
sas15kr1001-1 =197 MB/s
sas15kr1001-16 =209 MB/s
on several disk by running the 2 following commands
Code:
dd if=/dev/zero of=/sas15kr1001/test.io bs=64k count=1k conv=fdatasync
dd if=/dev/zero of=/sas15kr1001/test.io bs=64k count=16k conv=fdatasync
where sas15kr1001 is the mount point of a disk and "1" and "16" is the number thousand block of 64k written on the disk.
So i wrote this script
Code:
#!/bin/bash
adisk=( sas15kr1001 sas10kr501 sas15kr502 r10a r10b )
for i in "${adisk[@]}"
do
#       echo $i
        acount=( "1" "16" )
        for c in "${acount[@]}"
        do
                dd if=/dev/zero of=/`$i`/test.io bs=64k count=`$c`k conv=fdatasync 2>&1 |& awk '/copied/ {print "'$i'- '$c'=" $8 " "  $9}'
        done
done
While the command
Code:
dd if=/dev/zero of=/sas15kr1001/test.io bs=64k count=1k conv=fdatasync  2>&1 |& awk '/copied/ {print "sas15kr1001-1=" $8 " " $9}'
return
Code:
sas15kr1001-1=181 MB/s
when i run the script, the output is quite weird
Code:
./testio.sh
./testio.sh: line 9: sas15kr1001: command not found
./testio.sh: line 9: 1: command not found
sas15kr1001- 1=188 MB/s
./testio.sh: line 9: sas15kr1001: command not found
./testio.sh: line 9: 16: command not found
sas15kr1001- 16=200 MB/s
any one can point me where i get it wrong ?
Cheers

Last edited by limdel; 12-17-2016 at 05:51 AM.
 
Old 12-17-2016, 06:42 AM   #2
AwesomeMachine
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Registered: Jan 2005
Location: USA and Italy
Distribution: Debian testing/sid; OpenSuSE; Fedora; Mint
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I think you must change the tick marks to quotes. Ticks indicate the output of a command. `ls` is the output of ls.
 
1 members found this post helpful.
Old 12-17-2016, 07:21 AM   #3
MadeInGermany
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Registered: Dec 2011
Location: Simplicity
Posts: 1,294

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Yes, backticks run their contents as a subshell i.e. shell commands.
If not yet in quotes, quote the variables
Code:
of=/"$i"/test.io bs=64k count="$c"k
If they are in quotes already you might need to { } them
Code:
"of=/$i/test.io" bs=64k "count=${c}k"
--
Why array variables?
Code:
#!/bin/bash
for i in sas15kr1001 sas10kr501 sas15kr502 r10a r10b
do
#       echo $i
        for c in 1 16
        do
...
Array variables make sense if you want to do complex operations on the elements, like removing/inserting/rearranging/changing them.
 
1 members found this post helpful.
Old 12-18-2016, 09:33 PM   #4
limdel
Member
 
Registered: May 2009
Location: Ho Chi Minh City, Vietnam
Distribution: CentOS - FEDORA
Posts: 59

Original Poster
Rep: Reputation: 15
Thank you guys for your valuable feedback.
Here the final script for further ref.
Code:
#!/bin/bash
for i in sas15kr1001 sas10kr501 sas15kr502 r10a r10b
do
#       echo $i
	for c in 1 16
	do
		siz=`expr $c \* 64000`
		dd if=/dev/zero of=/"$i"/test.io bs=64k count="$c"k conv=fdatasync 2>&1 |& awk '/copied/ {print "'$i' \t '$siz' \t " $8 " "  $9}'
	done
done
 
Old 12-19-2016, 08:24 AM   #5
MadeInGermany
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Registered: Dec 2011
Location: Simplicity
Posts: 1,294

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Bash's builtin arithmetic is as good as the one in expr, but much more efficient.
Replace
Code:
		siz=`expr $c \* 64000`
by
Code:
		siz=$(( c * 64000 ))
 
  


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