shell - Why using variable in awk printf doesn't work? -

newbie here. trying use awk print few information. wrote shell scripts

#!/bin/bash  turbvar="u" bcname="outlet" str="$turbvar $bcname b.c. = " # method-1 awk -v rs='}' '/'$bcname'/ { printf "%20s = %s\n" $str $4; exit;}' bcfile  | tr -d ";" # method-2 awk -v rs='}' -v var1=$bcname '$0 ~ var1 { printf "%20s = %s\n" $str $4; exit;}' bcfile  | tr -d ";" 

the bcfile file contents are

boundary {     inlet     {         type            fixedvalue;         value           uniform (5 0 0);     }      outlet     {         type            inletoutlet;         inletvalue      $internalfield;         value           $internalfield;     }      .... }  

i hope output like

u outlet b.c. = inletoutlet 

sadly, not work. complains awk: (filename=0/u fnr=4) fatal: not enough arguments satisfy format string %20s = %s.

why can't use $str variable in awk printf?

second question, method better? using '/'$bcname'/ or using -v var1=$bcname '$0 ~ var1?, why cant use '/$bcname/ or '/"$bcname"/directly? difference between strong quote , weak quote here?

you code cleaned should be:

awk -v rs="}" -v v1="$bcname" -v s="$str" '$0~v1{gsub(/;/,"");printf "%s%s\n",s,$4;exit}'  u outlet b.c. = inletoutlet 

notes:

• don't play shell expansion , quoting it's real headache. pass in shell variables nicely -v.

• you need comma separator arguments printf.

• always quote shell variables!

• you should being doing gsub(/;/,"") inside awk script instead of tr -d ";".

however may not best approach couldn't no context provied.