PHP realpath('./models') returns "" or null?

Question

When I try

realpath('./')

I get "D:\Projects\Tickle\public"

But when I try

realpath('./models'); // or './models/'

I get "" or null. why is that? The strange thing is that when I try

realpath('../application/models');

I get "D:\Projects\Tickle\application\models" which is right

UPDATE

My directory structure looks like below (I am using Zend Framework)

/application
    /models
    /controllers
    /views
    /configs
    bootstrap.php  <-- this is where I am

Your current dir is not `applications` one? – Your Common Sense Dec 16 '10 at 13:41 — Your Common Sense, Dec 16 '10 at 13:41
What is your current directory? – Gumbo Dec 16 '10 at 13:42 — Gumbo, Dec 16 '10 at 13:42
See update. I am in /application/bootstrap.php – Jiew Meng Dec 16 '10 at 14:26 — Jiew Meng, Dec 16 '10 at 14:26

Nedec · Accepted Answer · 2010-12-16T16:48:38.687

3

According to the output you posted, your directory structure seems to look like this:

D:\Projects\Tickle
    public
        foo.php
    application
        models

We are in foo.php. By using realpath('./models'); you refer to D:\Projects\Tickle\public\models which does not exist.

UPDATE If you want to get a file (relative to the current file), you should use something like this:

realpath(__DIR__ . '/models');

Or even better (__FILE__ is the current file you are in):

realpath(dirname(__FILE__) . '/models')

Using the Zend Framework, the APPLICATION_PATH constant is defined as realpath(dirname(__FILE__) . '/../application') (in public/index.php).

edited Dec 16 '10 at 16:48

answered Dec 16 '10 at 13:43

Nedec

836
10
15

I missed the part where `realpath('./')` is public! but why? The code is actually in `/application/bootstrap.php` if you are familiar with Zend Framework. Isit because the code starts running from /public/index.php which calls bootstrap.php? using `realpath(__DIR__ . '/models')` works tho. Is `__DIR__` not same as `./`? – Jiew Meng Dec 16 '10 at 14:31
1

@jiewmeng You are in public since the file you call in the browser is the `public/index.php`. The bootstrap is just included in `public/index.php`, you are right. The difference between your two approaches is: `__DIR__` returns the directory of the current **file**. (It is a constant - different in each included file!) When you use `realpath('./')` you do not use the constant, the script takes `public/index.php` (called by the user) as starting point. Working with Zend, you should *always* use the constant `APPLICATION_PATH`, your models directory would be: `APPLICATION_PATH . '/models'`. – Nedec Dec 16 '10 at 16:44

score 2 · Answer 2 · answered Dec 16 '10 at 13:41

2

Try following and see if you get false by any chance:

var_dump(realpath('./models'));

If I'm not mistaken, your tree structure is as shown bellow and the file you have above code in is public, correct? In that case, you are actually trying to get a real path of public/models.

public
    somefile.php
application
    models

answered Dec 16 '10 at 13:41

David Kuridža

7,026
5
26
25

Yes I get false. Oh and I missed the part where `realpath('./')` is public! but why? The code is actually in `/application/bootstrap.php` if you are familiar with Zend Framework. Isit because the code starts running from /public/index.php which calls bootstrap.php? using `realpath(__DIR__ . '/models')` works tho. Is `__DIR__` not same as `./`? – Jiew Meng Dec 16 '10 at 14:29

PHP realpath('./models') returns "" or null?

2 Answers2