shell script error when declaring command

Question

Code below is supposed to check the memory for user and if its greater than 1000, print message I keep getting error- line 4: impala: command not found

#!/bin/bash

while [ true ] ;do
used= `ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" 
"int(a[i]/1024+0.5);}' | grep user`

if [[ $used > 1000 ]]; then
echo "user memory is $used"

fi
sleep 5
done

I have tried used= ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" "int(a[i]/1024+0.5);} | grep user'

and used= 'ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" "int(a[i]/1024+0.5);}' | grep user' I need a fresh eye on this. Please help.

remove the space

123
– 123

2017-06-27 16:09:32 +00:00
Commented Jun 27, 2017 at 16:09 — 123
– 123, Commented Jun 27, 2017 at 16:09
Remove the space after the equals sign?

perreal
– perreal

2017-06-27 16:10:52 +00:00
Commented Jun 27, 2017 at 16:10 — perreal
– perreal, Commented Jun 27, 2017 at 16:10
Jesus.. Thank you

TheNewGuy
– TheNewGuy

2017-06-27 16:10:56 +00:00
Commented Jun 27, 2017 at 16:10 — TheNewGuy
– TheNewGuy, Commented Jun 27, 2017 at 16:10

sjsam · Accepted Answer · 2017-06-27 17:12:32Z

1

In bash, as mentioned [ here ], putting spaces around the equal sign would cause errors, So the right format is

variable_name=value;

Moreover, you may change

while [ true ]

to

while true

Edit

If used has the form impala 600 and you're only interested in the number at the end, then you may do

used="${used##* }"
#Do this just after the your first command.

Finally do

#use -gt for integer comparisons and > for string comparisons
if ! [ -t $used ] && [ $used -gt 1000 ]
then
  echo "user memory is $used"
fi

^{Note: Though the syntax errors in the script is resolved there is no guarantee that the program logic is right}

edited Jun 27, 2017 at 17:12

answered Jun 27, 2017 at 16:47

sjsam

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2 Comments

TheNewGuy Over a year ago

my output from ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" "int(a[i]/1024+0.5);}' | grep impala is impala 600. Im trying to remove "impala" Ive tried ps hax -o rss,user | awk '{a[$2]+=$1;}END{for(i in a)print i" "int(a[i]/1024+0.5);}' | grep impala | sed 's/\D+/g' <<< "$s" with no luck

TheNewGuy Over a year ago

Thank you. another thing is im still getting output when I shouldnt. since 600 is not > 1000

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