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Remove arguments

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# 1  
Old 03-05-2009
Remove arguments
Hi all,
First of all, this is my first post in this forum, so it's nice to know you. I hope not to brake any 'rule' Smilie

I'd like to know, if it's possible to remove a parameter from command line, regardless of its position? I've found a solution using shift, if it's the first argument. But, I don't know in which position the argument comes (if it comes). Any solution?

Thanks a lot.

Albert.
# 2  
Old 03-05-2009
Assuming I understand -
let's say your arguments are 1 2 3 4
and you want to "lose" 2 (which in this case happens to be $2).
So the args become 1 3 4
Code:
# do not include the value you want to skip -- $2
set - $1 $3 $4
# show the new arguments
echo $*

# 3  
Old 03-05-2009
Quote:
Originally Posted by jim mcnamara
Assuming I understand -
let's say your arguments are 1 2 3 4
and you want to "lose" 2 (which in this case happens to be $2).
So the args become 1 3 4
Code:
# do not include the value you want to skip -- $2
set - $1 $3 $4
# show the new arguments
echo $*

Thanks jim, that could be a good solution. I tried something else that it doesn't work, using strings.
I think it could be easier to understand if I show you my code:
Code:
ARGS=$@

confirmed=false
for order in $ARGS; do
    if [ $order == "-confirmed" ]; then
        confirmed=true
    fi
    [....]
done

When I know "-confirmed" is an argument I have to remove it from ARGS.
I tried cut in this way:
Code:
ARGS=`echo $ARGS | cut -d $order -f-´

but the delimiter must be a character. I think using sed or awk should work, but I don't know how. I'm quite new in bash programming.

Thanks a lot.

Albert
# 4  
Old 03-05-2009
I got it:
Code:
ARGS=$@

confirmed=false
for order in $ARGS; do
    if [ $order == "-confirmed" ]; then
        confirmed=true
         ARGS=`echo "$ARGS" | awk 'BEGIN { FS="-confirmed" } ; { print $1$2 }'`
         continue
    fi
    [....]
done


But I have one more question. In awk command why it doesn't work if I replace
Code:
{ FS = "-confirmed" }

for
Code:
{ FS = "$order" }

It is supposed $order contains "-confirmed", isn't it?

Albert.

# 5  
Old 03-05-2009
Quote:
Originally Posted by AlbertGM
...
But I have one more question. In awk command why it doesn't work if I replace
Code:
{ FS = "-confirmed" }

for
Code:
{ FS = "$order" }

It is supposed $order contains "-confirmed", isn't it?

Albert.


Because awk has a special way of passing shell variables into it:

Code:
awk -v sep="$order"  'BEGIN { FS=sep }{ print $1 $2 }'

# 6  
Old 03-05-2009
ok you have to delete -confirmed so:
Code:
set - $( echo "$*" | sed 's/-confirmed//' )

Run this - you do not have to test for -confirmed if you do not need to
# 7  
Old 03-06-2009
Quote:
Originally Posted by rubin
Because awk has a special way of passing shell variables into it:

Code:
awk -v sep="$order"  'BEGIN { FS=sep }{ print $1 $2 }'

Quote:
Originally Posted by jim mcnamara
ok you have to delete -confirmed so:
Code:
set - $( echo "$*" | sed 's/-confirmed//' )

Run this - you do not have to test for -confirmed if you do not need to
Thanks a lot.

Albert.
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