Eliminate trailing and back slashes - escapes ?

[scanreg]

I found this for eliminating trailing and back slashes:

 

$clean = ereg_replace ("^[\/]+", "", $not_clean);

 

Doesn't the back slash need to be escaped, like this:

 

$clean = ereg_replace ("^[\\/]+", "", $not_clean);

 

Thanks : )

[Pikachu2000]

What happened when you tried it?

 

[cags]

1.) You should consider using preg_replace instead of ereg_replace as the latter is deprecated.

2.) Your pattern matches 1 or more capital V characters anchored to the start of the string and replaces them with noting.

3.) \V would mean any character that is not a vertical whitespace character. So essentially it would match anything that is not a newline character (making it very similar, if not the same as using .).

[scanreg]

It looks like a V but it's really a back slash and a forward slash:

 

\ /

 

I didn't want to put a space in there

[cags]

Oh, my bad. Well if the objective is to match any forward slashes at the start of the string you don't need either backslash as the forward slash character has no special meaning. The only reason to escape it is if you are using PCRE and have used it as a delimiter. But personally I'd suggest using a different delimiter, since the forward slash is all too common in patterns that involve paths or HTML.

[scanreg]

but what if I'm trying to match both forward slashes and back slashes:

 

$clean = ereg_replace ("^[\ /]+", "", $not_clean);  (added a space in between for clarity)

 

should it be like this:

 

$clean = ereg_replace ("^[\\ /]+", "", $not_clean);  (added space here too)

 

Thanks

[Psycho]

$clean = preg_replace("#^(\\\|/)*(.*?)(\\\|/)*$#", '${2}', $not_clean);

[scanreg]

$clean = preg_replace("#^(\\\|/)*(.*?)(\\\|/)*$#", '${2}', $not_clean);

 

Ah, you have to escape the escape for the escape to work, okay, got it

 

Thanks for the super nice preg_replace too, neat