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# 1  
Old 03-02-2012
perl Unix programming Question
Hello can somebody tell me whats wrong with this good

PHP Code:
 #/bin/perl -f
 substr( $_, -2, 1 ) = "" unless length( $_ ) % 2 
i get an error :
Code:
 ./ev: syntax error at line 2: `(' unexpected

please help

thank you
Last edited by radoulov; 03-03-2012 at 03:19 AM..
# 2  
Old 03-03-2012
Please post the entire code and explain what exactly you're trying to achieve.
# 3  
Old 03-03-2012
Quote:
Originally Posted by koricha
...can somebody tell me whats wrong with this good
PHP Code:
 #/bin/perl -f
 substr( $_, -2, 1 ) = "" unless length( $_ ) % 2 
i get an error :
Code:
 ./ev: syntax error at line 2: `(' unexpected

...
The shebang is incorrect, because the "!" character is missing after the "#".

The way it is right now, it's acting as a shell script comment. I say "shell script" because you are executing it as a shell script due to the way you've invoked it. And so the shell balks at the "(" character.

tyler_durden
# 4  
Old 03-03-2012
Quote:
Originally Posted by radoulov
Please post the entire code and explain what exactly you're trying to achieve.
Ok that is the cod i am trying to save in vi
steps

open vi example1

Code:
#!/bin/perl -f
   substl( $_, -2, 1 ) = "" unless length( $_ ) % 2

save and quite

Code:
chmod +x example1

Code:
./example1

and it gives me an error
Last edited by radoulov; 03-04-2012 at 04:43 AM.. Reason: Code tags, please!
# 5  
Old 03-03-2012
Hi koricha,

You are not helping us so much. Your code has no sense for me.

1. -f switch gives me next error: Can't emulate -f on #! line at ./example1 line 1
2. substl is not a valid function in perl.
3. $_ has no value.
4. Describe your error. We cannot see your screen.
5. You run your program without arguments. So, try to answer to radoulov and say what you want to achieve.

Regards,
Birei
# 6  
Old 03-03-2012
Never mind it works


PHP Code:
#!/bin/ksh
perl -pe 'substr( $_, -2, 1 ) = "" unless length( $_ ) % 2' 
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