"The use statement with non-compound name...." [duplicate] - php

Question

Getting this error when I put use Blog; at the top.
Warning: The use statement with non-compound name 'Blog' has no effect
in...
Blog is my namespace in which I have 3 classes: Article, List and Category and a few functions.
If I change my statememnt to use Blog\Article; then it works...
Can't I just specify the namespaces I want to use? Do I need to provide classes?
What if I have functions within that namespaces? When I call them outside of the namespace, I'm forced to prepend \Blog\ to each one's name...

Answer 1

PHP's use isn't the same as C++'s using namespace; it allows you to define an alias, not to "import" a namespace and thus henceforth omit the namespace qualifier altogether.
So, you could do:
use Blog\Article as BA;
... to shorten it, but you cannot get rid of it entirely.
Consequently, use Blog is useless, but I believe you could write:
use \ReallyLongNSName as RLNN;
Note that you must use a leading \ here to force the parser into knowing that ReallyLongNSName is fully-qualified. This isn't true for Blog\Article, which is obviously already a chain of namespaces:
Note that for namespaced names (fully qualified namespace names containing namespace separator, such as Foo\Bar as opposed to global names that do not, such as FooBar), the leading backslash is unnecessary and not recommended, as import names must be fully qualified, and are not processed relative to the current namespace.
http://php.net/manual/en/language.namespaces.importing.php

Answer 2

Since this question appears as the first result on Google for this error I will state how I fixed it.
Basically if you have a framework, say like Yii2 you will be used to having to do declare classes like:
use Yii;
use yii\db\WhatEver;
class AwesomeNewClass extends WhatEver
{
}
You will get this error on Use Yii since this class has no namespace.
Since this class has no namespace it automatically inherits the global symbol table and so does not need things like this defining, just remove it.

Answer 3

The use statement in PHP is really just a convenience to alias a long namespace into something that may be a little easier to read. It doesn't actually include any files or do anything else, that effects your development, besides providing convenience. Since, Blog isn't aliased as anything you aren't gaining any of the convenience. I could imagine you could do something like
use \Blog as B;
And that may even work. (It could be argued you actually lose convenience here by obscuring but that's not what the question is about) Because you're actually aliasing the Blog namespace to something else. Using Blog\Article works because, according to the docs:
// this is the same as use My\Full\NSname as NSname
use My\Full\NSname;
So your snippet would be equivalent to:
use Blog\Article as Article;

Answer 4

if you don't want to use 'as' syntax like
use \Blog as B;
define a namespace for the file
namespace anyname;
use Blog

Answer 5

The error "The use statement ... has no effect..." also pops up if you try to use a trait before a class definition.
use My_trait; // should not be here
class My_class{
// use My_trait; should be here instead
}

Answer 6

Perhaps it's something like
namespace Path\To\Your\Namespace\Blog;
use Blog; // Redundant
class Post {
public $linkedArticle;
public function __construct($article = null)
{
$this->linkedArticle = $article ?? new Blog\Article();
}
}
Blog is already available, because that's the namespace you're in, so you can use new Blog\Article(); without use Blog; at the top. That's exactly what the error tells you - the added line has no effect.
Pointless:
use SingleNonNestedClassThatIsAlreadyPresentInTheCurrentNamespace;
Useful:
use SingleNonNestedClassThatIsAlreadyPresentInTheCurrentNamespace as Phew;
If you on the other hand wish to use new Article() then you can do it like this.
namespace Path\To\Your\Namespace\Blog;
use Blog\Article; // Equivalent to "use Blog\Article as Article;"
class Post {
public $linkedArticle;
public function __construct($article = null)
{
$this->linkedArticle = $article ?? new Article();
}
}
In practice you'd do something like
// Fairly separated domains
use Some\TooLong\Namespace\App\User;
use Some\TooLong\Namespace\App\Ecommerce;
use Some\TooLong\Namespace\App\Auth;
but not necessarilly
// Two tools in same domain
use Some\TooLong\Namespace\App\Ecommerce\Cart;
use Some\TooLong\Namespace\App\Ecommerce\Checkout;
as well. I'm positive there are better examples than this ;)