Scripting with parameters

[carlbeech]

Hi

Here's an interesting one - and (hopefully!) simply answered!

I have a number of servers of different types (Linux, Solaris, AIX etc), each of which is running ksh (yes I *know* its better to use bash, however, this decision wasn't mine )

Anyway, I've a set of scripts, that ideally I'd like to keep the same on all servers, however, there are subtle differences between each flavour of unix. So my idea is to create a device specific 'base' API which gives the basic functions required by scripts, as such, the scripts can be kept the same....

All has worked well so far, until I came across a problem with AWK...

I thought I should be able to create a function:

API_awk()
{

awk $*

}

As the $* passes across all the parameters from the calling command line... however, I'm getting syntax errors etc...

e.g.
echo hello | awk '{printf $1}'
hello
- works ok...

echo hello | API_awk '{printf $1}'
Param: {printf $1}
Execute command: awk {printf $1}
awk: syntax error near line 1
awk: illegal statement near line 1

(I put some extra echos in to say what the script is doing...)
- As you can see it isn't working...

So.... my question is, a) what's going wrong? - I'm thinking this has something to do with substitution, and that the input is being piped in...

and b) how can this be made to work, in as similar way to a normal statement as possible?

Many thanks in advance...

Carl.

[drl]

Hi.

A quoting issue. This version seems to work:

#!/usr/bin/env ksh

# @(#) s0	Demonstrate avoiding quoting issues with awk.

# Section 1, setup, pre-solution.
# Infrastructure details, environment, commands for forum posts. 
# Uncomment export command to test script as external user.
# export PATH="/usr/local/bin:/usr/bin:/bin"
set +o nounset
pe() { for i;do printf "%s" "$i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
C=$HOME/bin/context && [ -f $C ] && . $C 
set -o nounset

API_awk()
{
set +o nounset

pe " debug API_awk, parameters :" "$*" ":"
awk "$*"

}

# Section 3, solution.
pl " Results:"
echo Hello world |
API_awk '
{print "Greetings from inside API_awk" }
{printf "%s,%s\n",$2,$1}
'

exit 0

producing:

% ./s0

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian GNU/Linux 5.0 (lenny) 
ksh 93s+

-----
 Results:
 debug API_awk, parameters :
{print "Greetings from inside API_awk" }
{printf "%s,%s\n",$2,$1}
:
Greetings from inside API_awk
world,Hello

Best wishes ... cheers, drl

[carlbeech]

Hi,

Many thanks for the reply - this works, however, when I try to use the following statements I get an error - can you let me know why?

ps -eaf|grep -v grep|grep pmon|API_awk '{print $NF}'

Outputs:
instance1
instance2
...

which is right, however:

ps -eaf|grep -v grep|grep pmon|API_awk '{print $NF}' | API_awk -F_ '{print $3}'

gives:
awk: syntax error near line 2
awk: bailing out near line 2

My definition of API_awk is:


API_awk()
{

awk "$*"

}


- it seems something to do with the -F parameter?

Thanks

Carl

[cfajohnson]

Perhaps:

API_awk()
{
  awk "$@"
}

[carlbeech]

Ok - this works!

- I've not met $@... whats the difference between $* and $@?

Thanks

Carl

[drl]

Hi.

...I've not met $@... whats the difference between $* and $@? ...

The  meaning  of $* and $@ is identical when not quoted or when
       used as a variable assignment value or as a file name.   However,  when
       used  as a command argument, "$*" is equivalent to "$1d$2d...", where d
       is the first character of the IFS variable, whereas "$@" is  equivalent
       to  "$1" "$2" ....

excerpt from man ksh, q.v.

Best wishes ... cheers, drl

[carlbeech]

Many thanks for the assistance - greatly appreciated!

Carl.