Using If-Statement to assign Variable in awk - awk

Question

I am trying to create an awk script file that takes an input file and converts the fourth column of information for the first three lines into a single row.
For example, if input.txt looks like this:
XX YY val1 1234
XX YY val2 2345
XX YY val3 3456
stuff random garbage junk extrajunk
useless 343059 random3
I want to print the fourth column for rows 1, 2 and 3 into a single row:
1234 2345 3456
I was trying to do this by using if/else-if statements so my file looks like this right now:
#!/usr/bin/awk -f
{
if ($1 == "XX" && $3 == "val1")
{
var1=$4;
}
else if ($1 == "XX" && $3 == "val2")
{
var2=$4;
}
else if ($1 == "XX" && $3 == "val3")
{
var3=$4;
}
}
END{
print var1,var2,var3
and then I would print the variables on one line. However, when I try to implement this, I get syntax errors pointing to the "=" symbol in the var2=$4 line.
EDIT
Solved, in my real file I had named the variables funky (yet descriptive) names and that was messing it all up. - Oops.
Thanks

Answer 1

you can write something like this
$ awk '$1=="XX"{if($3=="val1") var1=$4
else if($3=="val2") var2=$4
else if($3=="val3") var3=$4}
// .. do something with the vars ....
however, if you just want to print the fourth column of the first 3 lines
$ awk '{printf "%s ", $4} NR==3{exit}' ile
1234 2345 3456

Answer 2

Try this instead:
#!/bin/env bash
awk '
$1 == "XX" { var[$3] = $4 }
END { print var["val1"], var["val2"], var["val3"] }
' "$#"
There's almost certainly a much simpler solution depending on your real requirements though, e.g. maybe:
awk '
{ vars = (NR>1 ? vars OFS : "") $4 }
NR == 3 { print vars; exit }
' "$#"
. For ease of future enhancements if nothing else, don't call awk from a shebang, just call it explicitly.