使用fnmatch（）测试level1.level2.level3中的权限

Question

I am currently looking into re-working a permissions system in PHP that was built a while ago. The current version has exact permissions so for Admin you would have the following permissions:

acl.manage.self
acl.manage.others

That would say the user has access to manage his own permissions, as-well as others.

However while working on some new applications, I attempted to tap into the current permissions system by giving myself access to my entire application via:

lst.*

That way I don't have to give myself a bunch .view .manage etc.

It seems as though the existing system does not understand the * wildcard. So I am sitting here trying to re-write the "has_permission"

So my current working test environment I have created a user with the following permissions:

array(3) { [0]=> string(1) "*" [1]=> string(15) "fake.permission" [2]=> string(27) "none.of.these.should.matter" } 

Technically the only permission that matters is [0] which says I should have permission to do anything and everything I want.

I am at a loss as to how to implement this flow into my method:

function has_perm($perm){
    //if I am checking if I have the perm 'acl.manage.all'
    //This function should return true if I have any of the following:
    // *, acl.*, acl.manage.*, acl.manage.all   

    //exact check
    foreach($this->perms as $p){
        if($p===$perm){ return true; }
    }

}

any nudge in the right direction would be awesome.

Answer 1

Your current code returns true if the needed permission $perm exactly matches any of your permissions $p.

You just need to change it to return true if the needed permission $perm matches any of the fnmatch() patterns $p.

Try this:

foreach($this->perms as $p) {
    if (fnmatch($p, $perm)) { return true; }
}

You don't need to keep the exact match check, because fnmatch() will still detect exact matches if your permissions don't contain wildcards.