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Problem in bash script

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# 8  
Old 11-18-2010
Could also be written (in ksh and bash)
Code:
numOfParameters=$(($numOfParameters + 1))

# 9  
Old 11-18-2010
Quote:
Originally Posted by radoulov
Modified version:

Code:
neededParameters=2
numOfParameters=0
correctNum=0
while getopts "s:l:" opt
do
    case $opt in         # the quotes are unnecessary in this context
    s)
        serviceName=$OPTARG # no space after the equal sign
        numOfParameters=$(( $numOfParameters + 1 )) # I suppose you need arithmetic expansion here
        ;;
    l)
        fileName=$OPTARG #errorline 2 
        numOfParameters=$(( $numOfParameters + 1 )) 
        ;;
    ?)
        echo "unknown option"
        ;;
    esac
done

Thanks you, you solved my problem.
# 10  
Old 11-18-2010
Unfortunately I know nothing about bash (ksh user...) but know some posix...
As I just found a box (AIX 6.1) that seems to have bash I gave a try...
The only other syntax that seems to work is this one:
Code:
let numOfParameters="$numOfParameters + 1"

But again Im no bash specialist and dont know what the implementation I use is worth...
# 11  
Old 11-18-2010
Quote:
Originally Posted by programAngel
O.K.

But why do you think that?

Because from what I understand when there is colons after the letter then it mean that a parameter must come after it when you run the script.
I mean it mean that if you write in the shell:
script -s
then the colons after the s (in the script) will send error.
Hope I explain myself good.
Yep i just had a doubt that's why i removed my post just after ...
so i just went back to the RTFM part ...

Colon following letter means : the letter option requires an argument
An initial colon prevent getopts from printing an error message when invalid option are given
Smilie


---------- Post updated at 04:26 PM ---------- Previous update was at 04:23 PM ----------

Quote:
Originally Posted by vbe
Unfortunately I know nothing about bash (ksh user...) but know some posix...
As I just found a box (AIX 6.1) that seems to have bash I gave a try...
The only other syntax that seems to work is this one:
Code:
let numOfParameters="$numOfParameters + 1"

But again Im no bash specialist and dont know what the implementation I use is worth...
For lazy ppl like me :
Code:
let numOfParameters++

Smilie
# 12  
Old 11-18-2010
Thanks ctsgnb!
Could you explain how
Code:
let numOfParameters++

works?

I just tried also last posix way:
Code:
 numOfParameters=$((numOfParameters + 1))

Works also...
# 13  
Old 11-18-2010
Quote:
Originally Posted by vbe
Thanks ctsgnb!
Could you explain how
Code:
let numOfParameters++

works?

I just tried also last posix way:
Code:
 numOfParameters=$((numOfParameters + 1))

Works also...
Not sure about what you mean by "how it works?" but I suppose let set the var to 0 if not initialized or amibguous (not numeric or empty) and then increment it by 1 (a different step could be defined with the i+=n notation this would do i="$i + n") where n is the numeric step of course.

So all the following are equivalent the first one is just shorter to write (and maybe less parse consuming)
Code:
let numOfParameters++
let numOfParameters+=1
let numOfParameters="$numOfParameters + 1"

I tested the following on a linux machine and it works fine with the different sh ksh bash (i don't know if it behaves the same way on AIX) :

Code:
$ ps
  PID TTY          TIME CMD
  387 pts/67   00:00:00 bash
13882 pts/67   00:00:00 sh
14052 pts/67   00:00:00 ps
$ echo $$
13882
$ echo $i
1
$ let i++
$ echo $i
2
$ exec /bin/ksh
$ ps
  PID TTY          TIME CMD
  387 pts/67   00:00:00 bash
13882 pts/67   00:00:00 ksh
14225 pts/67   00:00:00 ps
$ echo $$
13882
$ echo $i

$ let i++
$ echo $i
1
$ exec /bin/bash
$ ps
  PID TTY          TIME CMD
  387 pts/67   00:00:00 bash
13882 pts/67   00:00:00 bash
14420 pts/67   00:00:00 ps
$ echo $$
13882
$ echo $i
/etc/profile.d/which-2.sh
$ i=
$ echo $i

$ let i++
$ echo $i
1
$

Last edited by ctsgnb; 11-18-2010 at 01:09 PM..
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