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Use Type as PropertyPart CustomType instead of only name #347

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Use Type as PropertyPart CustomType instead of only name #347

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d9621a8
Don't use only name for PropertyPart.CustomType
visubesy Jul 4, 2016
767aea3
Adapt tests to new behaviour of PropertyPart
visubesy Jul 4, 2016
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2 changes: 1 addition & 1 deletion ...tNHibernate.Testing/ConventionsTests/OverridingFluentInterface/PropertyConventionTests.cs
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Original file line number Diff line number Diff line change
Expand Up @@ -65,7 +65,7 @@ public void TypeShouldntBeOverwritten()

Convention(x => x.CustomType<int>());

VerifyModel(x => x.Type.Name.ShouldEqual(typeof(CustomUserType).AssemblyQualifiedName));
VerifyModel(x => x.Type.GetUnderlyingSystemType().ShouldEqual(typeof(CustomUserType)));
}

[Test]
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2 changes: 1 addition & 1 deletion src/FluentNHibernate.Testing/DomainModel/Mapping/ClassMapConventionsTester.cs
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Expand Up @@ -108,7 +108,7 @@ public void EnumsDontGetTypeOverriddenByConventionsIfExplicitlySet()
{
new MappingTester<MappedObject>()
.ForMapping(m => m.Map(x => x.Color).CustomType(typeof(int)))
.Element("class/property[@name='Color']").HasAttribute("type", typeof(int).Name);
.Element("class/property[@name='Color']").HasAttribute("type", typeof(int).AssemblyQualifiedName);
}
}
}
10 changes: 4 additions & 6 deletions src/FluentNHibernate/Mapping/PropertyPart.cs
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Expand Up @@ -183,7 +183,9 @@ public PropertyPart CustomType(Type type)
if (typeof(ICompositeUserType).IsAssignableFrom(type))
AddColumnsFromCompositeUserType(type);

return CustomType(TypeMapping.GetTypeString(type));
attributes.Set("Type", Layer.UserSupplied, new TypeReference(type));

return this;
}

/// <summary>
Expand All @@ -206,11 +208,7 @@ public PropertyPart CustomType(string type)
public PropertyPart CustomType(Func<Type, Type> typeFunc)
{
var type = typeFunc.Invoke(member.PropertyType);

if (typeof(ICompositeUserType).IsAssignableFrom(type))
AddColumnsFromCompositeUserType(type);

return CustomType(TypeMapping.GetTypeString(type));
return CustomType(type);
}

private void AddColumnsFromCompositeUserType(Type compositeUserType)
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