mapstruct 1.3.1FINAL: @Mapper(componentModel = "spring") doesn't support custom bean name references mapstruct/mapstruct#1427

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英文:

mapstruct 1.3.1FINAL: @Mapper(componentModel = "spring") doesn't support custom bean name references mapstruct/mapstruct#1427

问题

我的问题涉及到GitHub上此处的问题:https://github.com/mapstruct/mapstruct/issues/1427

我至少有两个同名版本的映射器。我想使用Spring的getBean/Autowired功能,但这在mapstruct的框架之外不起作用。 mapstruct 1.3.1FINAL: @Mapper(componentModel = "spring") doesn't support custom bean name references mapstruct/mapstruct#1427

我遵循了上面链接中提到的第二种解决方法:扩展Spring的bean命名策略。有人能够让这个善意的提议成功吗?

如果我按照那里的代码部分进行,bean的命名就不会生效。对我来说,为什么不生效是很清楚的:没有任何组件需要扫描,尤其是没有需要找到的组件。

如果我在映射器注释中添加componentModel = "spring",我会得到一个ConflictingBeanDefinitionException。不知道为什么。也许有一个尾部问题,类似于"猫在尾巴上"的问题?

英文:

My problem references to this issue here on github <https://github.com/mapstruct/mapstruct/issues/1427>

I got at least two versions of mappers with the same name. I want to use springs getBean/Autowired possibilities but this doesn't work out of the mapstructs box yet. mapstruct 1.3.1FINAL: @Mapper(componentModel = "spring") doesn't support custom bean name references mapstruct/mapstruct#1427

I followed the second workaround mentioned in the upper link: extend Springs bean naming strategy. Did someone ever get this well ment proposal working?

If i follow the code parts from there the bean naming doesn't take place. For me its's clear why not: there aren't any components to scan and especially to find.

If i add a componenModel = "spring" to the mapper annotation i get a ConflictingBeanDefinitionException. Don't know why. Maybe there's a cat in the tail problem?

答案1

得分: 2

按照Filip在这里的说明https://github.com/mapstruct/mapstruct/issues/1427,我遵循了他的方法,并进行了一些修改使其生效。我在链接中添加了解决方案的评论。

主要更改如下:

  1. 我在我的映射器中添加了componentModel = &quot;spring&quot;,并使用过滤器来排除在Spring Boot应用程序中实现了我所有映射器的接口(MapperInterface.class)的所有映射器类。

  2. 我在我的Spring Boot应用程序类中添加了:

    @ComponentScan(basePackages = { &quot;com.application.spring_boot_class&quot; }, excludeFilters = { @ComponentScan.Filter(value = { MapperInterface.class }, type = FilterType.ASSIGNABLE_TYPE) })

英文:

As stated from Filip here https://github.com/mapstruct/mapstruct/issues/1427 i followed his approach and with a few modifications it worked. I added a solution comment in the link.

The main changes are:

  1. i added componentModel = &quot;spring&quot; to my mappers and used a filter to exclude all of my mapper classes (the interface all of my mappers are implementing: MapperInterface.class) within the Spring Boot application.

  2. To my Spring Boot application class i added:

    @ComponentScan(basePackages = { &quot;com.application.spring_boot_class&quot; }, excludeFilters = { @ComponentScan.Filter(value = { MapperInterface.class }, type = FilterType.ASSIGNABLE_TYPE) })

答案2

得分: 1

我之前遇到过类似的问题,我是通过在一个配置类中使用Spring bean的定义来解决的,这个配置类是用@Configuration注解标注的,代码如下所示:

@Bean
public IMyMapper offerWebMapper() {
    return Mappers.getMapper(IMyMapper.class);
}

然后你可以使用@Autowired或者getBean来注入这个映射器。

英文:

I had this issue before, and I resolved it using the Spring bean definition in a configuration class, a class annotated with @Configuration, with the Mapstruct mapper call like below:

@Bean
public IMyMapper offerWebMapper() {
    return Mappers.getMapper(IMyMapper.class);
}

And then you can inject the mapper using @Autowired or getBean.

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  • 本文由 发表于 2020年9月11日 17:34:52
  • 转载请务必保留本文链接:https://go.coder-hub.com/63844432.html
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