How to match a string with an optional "/" at the end?

I am trying to match a URL such as;

http://www.testing.com/documents/dashboard

What I want is to match "/documents开发者_Python百科/dashboard" or "/documents/dashboard/". How can be this done?

preg_match('/^\/documents\/dashboard[\/|]$/i', $_SERVER['REQUEST_URI']);

This doesn't work? What should I enter after pipe (|) character in [] block to cover "nothing" as well?

[\/|]$ is wrong, because [] creates a character class. So what you are matching there is / or | followed by end of string. To do what you were thinking of:

preg_match('~^/documents/dashboard(/|$)$~', $_SERVER['REQUEST_URI']);

Although I think it's easier to use:

preg_match('~^/documents/dashboard/?$~', $_SERVER['REQUEST_URI']);

/? means match the / character 1 or 0 times.

Tip: If you use a delimiter other than /, you won't have to escape forward slashes in the pattern.

preg_match('/^\/documents\/dashboard\/?$/i', $_SERVER['REQUEST_URI']);

People often use it at the end of their regex, but it can in fact be used anywhere. Although in your case it's just simpler to use the ? quantifier.

Or, if you absolutely want to use $, the following would work too:

/^\/documents\/dashboard(\/|$)$/i