spring/scala:可以自动装配 bean 依赖项吗?
我正在尝试在 spring(在 scala 中)没有 xml 配置的情况下执行以下操作
I'm trying do the following without xml configuration in spring (in scala)
<beans ... >
## (1)
## Auto create a bean by classname
## Auto-wires properties of x.y.Bar
<bean id="baz" class="x.y.Baz"/>
## (2)
## Create a x.y.Foo and auto-wire the property
<bean id="foo" class="x.y.Foo">
<property name="b" ref="baz"/>
</bean>
</beans>
我们有:
class Baz {}
class Foo {
@Autowired //?
@BeanProperty
val baz:Baz = null
}
我有以下测试设置:
@Configuration
class Config {
//@Autowired // Seem not to help
@Bean //( autowire=Array( classOf[Autowire.BY_TYPE ]) )
def foo: Foo = null
}
class BeanWithAutowiredPropertiesTest {
@Test
@throws[Exception] def beanWithAutowiredPropertiesTest(): Unit = {
var ctx = new AnnotationConfigApplicationContext(classOf[Config]);
val foo = ctx.getBean(classOf[Foo])
assertTrue(foo != null)
assertTrue(ctx.getBean(classOf[Foo]).baz != null)
}
}
我了解几个简单的替代方案:
I understand a couple of simple alternatives:
@ComponentScan -- 这种方法有几个问题:
@ComponentScan -- this approach has several issues:
- 不精确 - 一个包中可以有许多与自动连接类型匹配的类
- 它(本身)不允许为属性选择特定值
- 大型项目的扫描速度非常慢
无法将@Component 添加到 3rd 方类
- imprecision - there can be many classes matching an auto-wired type in a package
- it doesn't (in itself) permit selecting specific values for properties
- scanning is painfully slow for large projects
Can't add @Component to 3rd party classes
(如果我可以按名称注册候选自动连线类型,那会有很大帮助!)
(If I could register candidate auto-wire types, by name, that would help a lot!)
.
@Bean
def foo:Foo = {
val f = new Foo()
f.baz = ?? grrr! where from? Not available in this Config
f
}
然而:
这有点绕开了自动接线的问题.如果我明确选择了要设置的参数 baz,那么我首先需要实际获得对它的引用才能做到这一点.通常这可能很困难,特别是如果要使用的实际 baz 可能在另一个 @Configuration 中指定.
this sorta circumvents the point of auto-wiring. If I explicitly chose a parameter, baz to set, then I need to actually get a reference to it to do that in the first place. Frequently this can be difficult, especially if the actual baz to be used might be specified in another @Configuration.
因为我正在创建对象,所以我不需要自动连接它的所有依赖项吗?如果 Baz 有 100 个属性,而我只能明确指定 1 个属性并自动连接其余属性怎么办?
because I'm creating the object, don't I need to auto-wiring all of its dependencies? What if Baz has 100 properties and I only way to specify 1 explicitly and have the rest auto-wired?
AFAIK,基于 xml 的配置没有任何这些问题 - 但我不知所措,因为 spring 手册说你可以通过注释做所有相同的事情.
AFAIK, the xml based configuration doesn't have any of these problems - but I'm at a loss because the spring manual says you can do all the same things via annotations.
注意.我也看到:
@Bean( autowire=Array( classOf[Autowire.BY_TYPE ]) )
也许是可能的.我在网上找不到示例,scala 抱怨(注释参数不是常量).
might be possible. I can't find example online, and scala complains (annotation parameter is not a constant).
[已编辑]
class ApplicationController @Inject() (messagesApi: MessagesApi){
... code here ...
}
messagesApi 是一个注入成员
messagesApi is an injected member
由 Joshua.Suereth 回答.这是简短版本(丑陋但有效"):
Answered by Joshua.Suereth. This is the short version ("ugly but works"):
var service: Service = _;
@Autowired def setService(service: Service) = this.service = service