PHP use namespace - A solution to including all classes under a namespace

Question

Is there a solution to having to include all classes individually under a namespace?

My Laravel files are getting huge because I have to keep including a heck load of namespaces... It's pretty awful!

As a temporary solution, why might the following not be working:

namespace.Blah.txt:

use Blah\Blah; 
use Blah\Bloh;

php code:

eval( file_get_contents( "namespace.Blah.txt" );

If I could get this to work, I could evaluate the contents of a file... I do understand it's a bit noob... but... dammit!

I have about 20 classes at the moment. I pretty much have to include all of them in every controller... I'm really not that fussed about declaring memory for the classes as they are TINY... and they are never auto instantiated so will never be filled. This just isn't a problem for me at this moment in time... If it becomes one, I will go back to the old way... But I'm not understanding why the eval() is not working? — Jimmyt1988
– Jimmyt1988, Commented Apr 29, 2015 at 14:20

Antonio Carlos Ribeiro · Accepted Answer · 2015-04-29 14:26:09Z

7

There isn't, but in PHP 7 you'll be able to

use FooLibrary\Bar\Baz\{ ClassA, ClassB, ClassC, ClassD as Fizbo };

As the following RFC has passed:

https://wiki.php.net/rfc/group_use_declarations

EDIT:

Note that too many uses in a class may be a sign of "smell". Isn't that particular class doing too much? Shouldn't you be creating new "base" classes and extending them?

edited Apr 29, 2015 at 14:26

answered Apr 29, 2015 at 13:56

Antonio Carlos Ribeiro

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7 Comments

Jimmyt1988 Over a year ago

Thanks for that, I wondered if you might then know if I can do the above (eval)... I know it's horrid... But i'm desperate! And it can be changed at some point if it gets to be noob!

Antonio Carlos Ribeiro Over a year ago

Too many uses in a class may be a sign of "smell". Isn't that particular class doing too much? Shouldn't you be creating new "base" classes and extending them?

Antonio Carlos Ribeiro Over a year ago

I'm pretty sure eval will not work in this case, because uses are processed by the compiler. It's only a hint for the compiler to know where to find the class, so if you execute the use from elsewhere it will not work on your current class.

Jimmyt1988 Over a year ago

Hah. Thanks for the tip man. Any ideas why you can't evaluate the use keyword?

Antonio Carlos Ribeiro Over a year ago

Because it's not really for evaluation, it's just a hint. You can replace use Foo; Foo::whatever() by \Foo::whatever();, so it's something only to tell PHP where to find that class.

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