linux - How to loop over shell script, with different variable, store entire contents of script with variable, and then run another function -

suppose have script test.sh contains

#!/bin/bash  var in var1 var2;   in `seq 2`;     sh test2.sh $var > tmp.sh     cat tmp.sh;   done done 

and have script test2.sh looks such

#!/bin/bash  echo "i use variable here $1" echo "and again here $1" echo "even third time here $1" 

now in first script want pass entire content of test2.sh current local variable (i.e. on line six: sh test2.sh $var > tmp.sh) if call sh test2.sh var1 return

i use variable here var1 , again here var1 third time here var1 

so want pass entire content of sh test2.sh $var new shell file, argument in place of variable. hence output should be:

i use variable here var1 , again here var1 third time here var1  use variable here var1 , again here var1 third time here var1  use variable here var2 , again here var2 third time here var2  use variable here var2 , again here var2 third time here var2 

thus wondering is; how pass entire shell local argument, new, temporary shell script? wondering how run this:

for var in var1 var2;   in `seq 2`;   sh (sh test2.sh $var)   done done 

thanks.

you can read contents of second script test2.sh , execute in test1.sh arguments replaced variable value, this:

#!/bin/bash  var in var1 var2;     in `seq 2`;         # contents of test2 variable         script=$(cat test2.sh)         # set arguments of script in variable , execute         eval "set -- $var; $script"     done done 

but, read on risks of using eval, e.g. here: why should eval avoided in bash, , should use instead?