Using positional arguments in a bash script

Question

This is my script script.sh:

Numbr_Parms=$#
a=`expr $Numbr_Parms - 2`
while [ $a -le $Numbr_Parms ]
do
    if [ "$a" =  "3" ]
    then
       PARAMSTRING="-param $3"
    else
       PARAMSTRING="$PARAMSTRING -param $a"
    fi
    a=`expr $a + 1`
done
echo $PARAMSTRING

Running:

script.sh  username pass p1=v1 p2=v2 p3=v3

Expected output:

-param p1=v1 -param p2=v2 -param p3=v3

But i am getting: $PARAMSTRING as

-param p1=v1 -param 4 -param 5

Not sure what is the Issue with $4 and $5

Would you consider rewriting your script by using getopt and handling your options and parameters in a slightly different way to take advantage of getopt ? — louigi600
– louigi600, Commented May 26, 2016 at 12:23

chepner · Accepted Answer · 2016-05-26 12:51:04Z

2

If you want to skip the first two positional parameters, just use

for arg in "${@:3}"; do
    PARAMSTRING+="-param $arg "
done

The right way to build up a sequence of parameters, though, is to use an array, which will work even if one of the arguments itself contains whitespace.

for arg in "${@:3}"; do
    PARAMS+=(-param "$arg")
done

mycommand "${PARAMS[@]}"

edited May 26, 2016 at 12:51

answered May 26, 2016 at 12:25

chepner

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3 Comments

fedorqui Over a year ago

You may want to add a trailing space; otherwise, strings will overlap.

chepner Over a year ago

True. Actually, that also makes me wonder why I'm not using an array like I should be.

chepner Over a year ago

Try this: set a b c d e; echo "${@:3}".

user5076313 · Accepted Answer · 2016-05-27 03:37:09Z

The problem is in the below line.

PARAMSTRING="$PARAMSTRING -param $a"

In this the $a does not means the positional argument. It is a variable it have the 4, 5, like that only.

If you print that variable you will not get the positional argument. You get only 4, 5, 6 only.

So, you have to get the positional argument using that variable a.

Try the below step it will work as you expect.

Numbr_Parms=$#
a=`expr $Numbr_Parms - 2`
while [ $a -le $Numbr_Parms ]
do
    if [ "$a" =  "3" ]
    then
        PARAMSTRING="-param $3"
    else
        eval pos=\$$a    # This is the indirect reference to get the positional argument from variable a.
        PARAMSTRING="$PARAMSTRING -param $pos"
    fi
    a=`expr $a + 1`
done

echo $PARAMSTRING

Community · Accepted Answer · 2017-05-23 12:22:36Z

0

In:

PARAMSTRING="$PARAMSTRING -param $a"

You want to append the parameter in the position a. However, $a just holds the number. So to access $1, $2, etc you have to use variable indirection:

PARAMSTRING="$PARAMSTRING -param ${!a}"

The full script can be reduced to a simple loop over the parameters, using $# as limit:

for ((i=3; i<=$#; i++)); do
        PARAMSTRING="$PARAMSTRING -param ${!i}"
done
echo "$PARAMSTRING"

Or, even better, you can use an array like chepner shows in his answer.

edited May 23, 2017 at 12:22

CommunityBot

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answered May 26, 2016 at 11:37

fedorqui

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