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# 1  
Old 09-11-2008
Need help with grep command

Hi guys, i wondering with any1 could help me with this.
I have this script will i need to "grep" an integer form a file x2. If the value is 0 then it's ok. But anything greater then 0, i'll ask the script to echo an error message. This is how the script looks like:
Code:
MESSAGEFILE=x2
STRING="0"

if  grep "$STRING"  /tmp/$MESSAGEFILE > /dev/null
 then
  echo " data is running smoothly"
  exit 0
fi
echo " - Data is pilling up"
exit 2

the problem i'm having now is how do i grep an integer and check whether if it's greater then or equals to 0?
Any advice would be greatly appreciated.

Last edited by hanyou.lin; 09-11-2008 at 04:59 AM..
# 2  
Old 09-11-2008
How does the file x2 looks like? And how to handle if you have more lines with integers?
Post the contents of the file and the desired output.

Regards
# 3  
Old 09-11-2008
oh sorry. ok the x2 file looks like this.

count
--------
0


my desired output is for the script to grep the 3rd line which is an integer and check whether if its bigger then 0. Then print out the respective massage. i've already done the script for printing out the massage. But i don't know how to grep the integer for x2 and check if it's bigger then 0 or not.
Thanks alot
# 4  
Old 09-11-2008
It should be something like:

Code:
op=`awk 'NR==3{print}' x2`
if [ "$op" -gt 0 ]; then
  echo "Output is greater then 0"
fi

Regards
# 5  
Old 09-11-2008
Code:
op=`awk 'NR==3{print}' x`

if ["$op" -gt 0]; then
  echo "data is running smoothly"
  exit 0
fi

echo "Data Awaiting Upload!"

exit 2

So this is how the whole thing looks like. But im having error. it says
line 3: [ 0: command not found
# 6  
Old 09-11-2008
You need a space after the [

Might as well use awk all the way, though.

Code:
if awk 'NR==3{ if ($0) exit 0; exit 1 } x'; then
  echo data is running smoothly
  exit 0
fi

 

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