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Help with parsing parameters

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# 1  
Old 06-06-2006
Help with parsing parameters
Hi:- I need to parse a script 3 parameters (file, subject and email address). This is what I currently have:

allargs=$*
argcount=`echo $allargs | awk -F: '{ print NF }' ` # Total Number of arguments
pdffile=`echo $allargs | awk -F: '{ print $1 }' ` # PDF/binary file to be encoded
subject=`echo $allargs | awk -F: '{ print $2 }' ` # Parsed Subject field
reciplist=`echo $allargs | awk -F: '{ print $3}' ` # Parsed multiple/single smtp recipientlist
#outfile="`${basenam} $pdffile|sed s/out/pdf/`" # $pdffile without the path

It then does a check to make sure there are 3 argument.

This is how I am running the script:
./email_report.sh pdf_file subject email_address

but for some reason it thinks that there is only 1 argument and then it dies. But it does pickup the pdffile, subject and reciplist (as it echo's these as part of the check)

Anyone able to assist as to why it picks up the 3 arguments as 1?

Thanks
# 2  
Old 06-06-2006
couldnt get
Actually, you can refer to the parameters passed to script as $1,$2,$3 and so on.... I am not sure why you are using awk inside the script.
For a script executed as
Code:
myscript.sh first second third

the arguments are referred to inside the script as
Code:
arg1=$1  #arg1=first
arg2=$2  #arg2=second
arg3=$3  #arg3=third

Your script may be considering the 3 arguments as one because of the delimiter given to awk.
# 3  
Old 06-06-2006
Why do you do that?

if [ $# = 3 ]; then
arg1=$1
arg2=$2
arg3=$3
else
exit 1
fi
...

Perhaps what you want is to treat the last "argumet" as a (possible) list, in that case:

if [ $# > 3 ]; then
arg1=$1
shift
arg2=$1
shift
arg3=$*
else
exit 1
fi

...
# 4  
Old 06-06-2006
yes
Quote:
Perhaps what you want is to treat the last "argumet" as a (possible) list, in that case:
That could be a possibility!!!
# 5  
Old 06-07-2006
Quote:
Originally Posted by janet
allargs=$*
[snip]
Anyone able to assist as to why it picks up the 3 arguments as 1?
It could be because of the behaviour of $*. You should try using $@ instead.

Quoting from man sh

Code:
       *      Expands to the positional parameters, starting from  one.   When
              the  expansion occurs within double quotes, it expands to a sin-
              gle word with the value of each parameter separated by the first
              character of the IFS special variable.  That is, "$*" is equiva-
              lent to "$1c$2c...", where c is the first character of the value
              of  the IFS variable.  If IFS is unset, the parameters are sepa-
              rated by spaces.  If IFS is  null,  the  parameters  are  joined
              without intervening separators.
       @      Expands  to  the positional parameters, starting from one.  When
              the  expansion  occurs  within  double  quotes,  each  parameter
              expands to a separate word.  That is, "$@" is equivalent to "$1"
              "$2" ...  When there are no positional parameters, "$@"  and  $@
              expand to nothing (i.e., they are removed).

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