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{dir.name(String $url, Int $levels=1)}
{dir.name()}
— Returns a parent directory's path
Given a string containing the path of a file or directory, this function will return the parent directory's path that is levels up from the current directory.
Note:
{dir.name()} operates naively on the input string, and is not aware of the actual filesystem, or path components such as "..".
Returns the path of a parent directory as a
string
.
If there are no slashes in path, an empty string is returned.
Otherwise, the returned string is path with a trailing /component.
Last modified: 2021-06-07
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