Do ... Loop

Shell Programming and Scripting
1

ill be the first to admit i am a bit of a bash scripting noob... but ive come againest a small problem...

LOG=/var/log/local2.log

for i in SEC-0 SEC-1 SEC-2 SEC-3 SEC-4 SEC-5 SEC-6 SEC-7
do
grep $i $LOG | wc -l
done

The above works and gives me 8 different values. Later on in the script I wish to echo these strings, so i would like them as varibles.
Is there anyway to assign each total against a varible so when this is finished I have... something like..
$SEC-0=some number
$SEC-1=some number ..... etc etc

I hope I have made this clear, thanks in advanced....

2

Not really, but you can either use a more parowerful scripting language and use arrays, or create a variable containing your 'SEC-?' entries.

#!/bin/sh
LOG=/var/log/local2.log
SEC="SEC-0 SEC-1 SEC-2 SEC-3 SEC-4 SEC-5 SEC-6 SEC-7"
for i in $SEC
do
  grep $i $LOG | wc -l
done

echo "$SEC" | awk '{ print $3 }'   # prints "SEC-2" (the third 'SEC' string)
grep `echo "$SEC" | awk '{ print $5 }'` $LOG | wc -l   #searches the log for "SEC-6"

Does that help?

3

Ive now found what I was looking for ...

LOG="/var/log/local
level=(Emergency Alert Critical Error Warning Notification Informational Debugging)
for i in 0 1 2 3 4 5 6 7
do
        printf "${level[$i]} - " ;  grep -c SEC-$i $LOG
done

But the only thing im unsure of is how to allow for the wildcard in for the $LOG variable ????

edit by bakunin: please use "code"-tags when you post program code. Thx.

4

If you don't use quotes the wildcards should be evaluated.

LOG=/var/log/local?.log
5

it doesnt seem to like that it says ..

./sysloghtml.sh: line 14: [: /var/log/local2.log: binary operator expected
Unable to find logfile at /var/log/local?.log

6

What is in line 14 of your script?

If you are testing against a wildcard, you will get an error if it expands to more than one filename.