ksh escape a character

Shell Programming and Scripting
1

friends,

I have a situation where i am using a $RANDOM function along with the filename, I want this to be escaped by the OS in the first assignment (works as expected) and executed in the second assignment (does not execute $RANDOM)

filename1=filename1_\$RANDOM
echo $filename1
filename2=filename1
echo $filename2

The output of $filename1 is filename1_$RANDOM
The output of $filename2 is filename1_$RANDOM (i am expecting the $RANDOM to be substituted by a random number in the second assignment, since there is no escape character not sure why the shell is not substituting $RANDOM with a number)

2

Try

filename2=$(eval echo $filename1)
1 Like
3

Why do you need dynamic variable names? Time and time again we get asked for this, and there's extremely rarely any good reason to do this. It's insecure and error-prone.

4

Thanks, It works perfect !