AWK: parsing arguments in a loop - bash

Question

I'm trying to write a simple script that will display the fields specified by the user as bash arguments. For example I've got text file looks like this:
1 2 3 4 5
1 2 3 4 5
a b c d e
And for example user types:
./script.sh text 1 2 5
Where $1 = text, and other parameters (like $2 $3 and $4) are the fields, so output will look like this:
1 2 5
1 2 5
a b e
I've got this code which prints all the columns defined as a arguments, but one below the others:
#!/bin/bash
text="$1"
shift
for x in $#; do
awk '{print $var}' var="$x" $text
done
Output for example ./script.sh text 1 2 5:
1
1
a
2
2
b
5
5
e
I guess output looks like that because loop "for" is outside of AWK. Is it a good solution for this task to place the loop inside AWK? I tried a few things but always have trouble with the syntax.
Thank you for your time and help!

Answer 1

file="$1"
shift
awk -v flds="$*" 'BEGIN{n=split(flds,f)} {for (i=1;i<=n;i++) printf "%s%s", $(f[i]), (i<n?OFS:ORS)}' "$file"

Answer 2

You don't need to loop over the params, pass all of them to awk with -v option:
awk -v v1=$2 -v v2=$3 -v v3=$4 '{print $v1, $v2, $v3;}' $1
You may want to perform additional checks such as whether the file ($1) contains enough fields, the file ($1) exists etc. But the idea is the same.
In your code, you are reading the file multiple times, each checking for only a particular field but to get your desired output, each line must be checked for multiple fields at the same time.
Pass the columns to awk and split them into an array and print the column corresponding to each value in the array:
file=$1
shift
p="$#"
awk -v l="$p" '{t=split(l,a," "); for (i=1;i<=t;i++) printf $(a[i]) " ";printf "\n";}' $file