Or one a a myriad othersmy $fileName = "fileLocn/".$var1."_$var2.txt"; my $fileName = "fileLocn/${var1}_$var2.txt"; my $fileName = sprintf("fileLocn/%s_%s.txt", $var1, $var2);
In reply to Re: Use scalars to define a filename
by Utilitarian
in thread Use scalars to define a filename
by akrrs7
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |