I want to use regular expression to remove string with $ , % , # these three characters , but it seems can't remove $ and the error information shows undefined variable
How can I solve this problem?
here is my code
perl Remove.pl $ABC#60%
#!/usr/bin/perl
$Input = $ARGV[0];
$Input =~ s/\$|%|#//g;
print $Input;
thanks
I think your problem is with the shell, not with the Perl code. Single quote the argument to the script:
perl remove.pl '$ABC#60%'
The shell can interpret '$ABC' as a variable name in which case the script will receive no arguments. Perl will then complain about undefined variable in substitution.
$Input =~ s/[\$%#]//g;
ought to work
if you just want to remove some charactor, it will be better use tr
try this:
perl -e '$arg = shift; $arg =~ tr/$%#//d; print $arg' '$asdf#$'
your code is just fine, but the parameter you pass to the program will expand in bash. you should put single quote.
try this:
perl Remove.pl '$ABC#60%'
