I have a dev org, in which I have got a namespace prefix. In apex, I want to get this namespace prefix as string. Is there any straight way or I have to describe some object?

  • Unless I am missing something , this will work for managed package because MyClass will already have a namespace (our namespace). So that query will always return our namespace and not the org namespace where the app is installed. Commented Dec 20, 2014 at 14:05

3 Answers 3


Query for it, using a class that you know will be there:

ApexClass myClass = [
    select NamespacePrefix
    from ApexClass
    where Name = 'SomeClassNameHere'
  • Does this require any permissions that ordinary users may not have?
    – kresho
    Commented May 4, 2016 at 12:45
  • Hm good question. I think it should just work for all users when queried from Apex, but you'd want to test that to be sure.
    – jkraybill
    Commented May 7, 2016 at 5:10

Spotted in the Spring `16 release notes under Changed Objects:


The following new fields have been added:

  • NamespacePrefix
  • SignupCountryIsoCode—Represents the two-character ISO country code specified by the user for a sign-up request.

It appears it will now be possible to directly query for the NamespacePrefix without needing to have a known Apex Class or Trigger present.


Here's a technique to get the namespace without invoking SOQL.

// PageReference behavior is different between managed and unmanaged code.
// In managed code it modifies the URL to add the namespace to the page URL,
// so /apex/page becomes /apex/MY_NAMESPACE__page.
String namespaceOfMyManagedPackage =
    new PageReference('/apex/page')
    .replaceFirst('^/apex/', '')
    .replaceFirst('page$', '')
    .replaceFirst('__$', '');
// For example, if you are in a managed package (or the packaging org) this 
// will return 'MY_NAMESPACE', but if this is in unmanaged code it will
// be a blank/empty string


Here is a less hacky way to get the namespace. Extract it directly from the Type name:

// Experiment in Execute Anonymous Apex
String namespaceOfClass = AssertException.class.getName().substringBefore('AssertException').split('\\.')[0];
System.debug('CORRECT: ' + AssertException.class + '->' + namespaceOfClass);
//Output: 'CORRECT: System.AssertException->System'

Note: When you are extracting the namespace from inner types, to correctly get the namespace you must pass the top level class name to substringBefore.

Here are the experiments to verify it for yourself:

// To run the following experiments, save a New Apex Class 'Foo' in your org
    public class Foo { public class InsideFoo {} }

// Experiments in Execute Anonymous Apex
String namespaceOfFoo = Foo.class.getName().substringBefore('Foo').split('\\.')[0];
System.debug('CORRECT: ' + Foo.class + '->' + namespaceOfFoo);
//Output if Foo is in no-namespace class: 'CORRECT: Foo->'
//Output if Foo is in 'mynamespace' managed package: 'CORRECT: mynamespace.Foo->mynamespace'

String namespaceOfFooFromInsideFoo = Foo.InsideFoo.class.getName().substringBefore('Foo').split('\\.')[0];
System.debug('CORRECT: ' + Foo.InsideFoo.class + '->' + namespaceOfFooFromInsideFoo);
//Output if Foo is in no-namespace class: 'CORRECT: Foo.InsideFoo->'
//Output if Foo is in 'mynamespace' managed package: 'CORRECT: mynamespace.Foo.InsideFoo->mynamespace'

// Note that substringBefore needs the top level class name, NOT the inner class, otherwise
// it will not properly determine the class's namespace.
String MAYBETheNamespaceOfFoo = Foo.InsideFoo.class.getName().substringBefore('InsideFoo').split('\\.')[0];
System.debug('MAYBE INCORRECT: '+Foo.InsideFoo.class + '->' + MAYBETheNamespaceOfFoo);
//Output if Foo is in no-namespace class: 'MAYBE INCORRECT: Foo.InsideFoo->Foo'
//Output if Foo is in 'mynamespace' managed package: 'MAYBE INCORRECT: mynamespace.Foo.InsideFoo->mynamespace'
  • Sort of hacky but works!
    – Raul
    Commented Sep 12, 2017 at 10:33

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