Question

我正在寻找一种方法来操作初始化时定义的@ConfigurationProperties，这样当@Inject / @Autorwired使用@Bean方法配置的对象时，它就会被正确配置。

情景：

我在.yml文件中设置了一堆属性。因为所有这些属性都与字符串匹配，所以它们中的一些需要特殊处理才能正确配置为各自的类型（其中一些是枚举）。我是否将我的属性对象设为@Component，然后将其注入@Bean方法并进行修改？我尝试组合@Bean和@ConfigurationProperties注释，但是在@Bean方法本身返回之后配置对象，因此任何操作都将丢失/不可能。这样做的最佳方式是什么？

实施例：在我的.yml我有这个：

properties:
  sports:
    "football": ["Team 1", "Team 2", "Team 3"]
    "basketball": ["Team 4", "Team 5", "Team 6"]
  settings:
    "football":
      "period1": "45"
      "period2": "90"
      "players": "11"
    "basketball":
      "periods": "4"
      "players": "5"

这些与以下对象匹配：

Map<SportsEnum, List<TeamsEnum>
Map<SportsEnum, Map<SportSettingsEnum, String>>

TL; DR：我想从.yml/.properties文件中获取一个对象并操纵其可注入表示。请提供具体的例子！

Answer 1

这肯定是支持的，我的猜测是你做错了什么。让我们一步一步走。

首先，您的YAML文件格式错误，它应该如下所示：

properties:
  sports:
    football:
      - Team1
      - Team2
      - Team3
    basketball:
      - Team4
      - Team5
      - Team6
  settings:
    football:
      period1: 45
      period2: 90
      players: 11
    basketball:
      periods: 4
      players: 5

然后，您的配置属性如下所示：

@ConfigurationProperties(prefix = "properties", locations = "classpath:sports.yml")
public class SportsProperties {

    private Map<SportsEnum, List<TeamsEnum>> sports;
    private Map<SportsEnum, Map<SportSettingsEnum, String>> settings;

    public Map<SportsEnum, List<TeamsEnum>> getSports() {
        return sports;
    }

    public void setSports(Map<SportsEnum, List<TeamsEnum>> sports) {
        this.sports = sports;
    }

    public Map<SportsEnum, Map<SportSettingsEnum, String>> getSettings() {
        return settings;
    }

    public void setSettings(
        Map<SportsEnum, Map<SportSettingsEnum, String>> settings) {
        this.settings = settings;
    }

    @PostConstruct
    public void customManipulation() {
        System.out.println(sports);
        System.out.println(settings);
    }
}

接下来，应在配置中启用属性：

@SpringBootApplication
@EnableConfigurationProperties(SportsProperties.class)
public class YamlSampleApplication {

    public static void main(String[] args) {
        SpringApplication.run(YamlSampleApplication.class, args);
    }

}

应该是这样，真的。这是我的枚举：

public enum SportsEnum {
    football, basketball
}

public enum TeamsEnum {
    Team1, Team2, Team3, Team4, Team5, Team6
}

public enum SportSettingsEnum {
    periods, period1, period2, players
}

这是我在日志中得到的输出：

2016-07-26 17:44:41.226 DEBUG 30015 --- [           main] s.b.e.YamlPropertySourceLoader$Processor : Loading from YAML: class path resource [sports.yml]
2016-07-26 17:44:41.282 DEBUG 30015 --- [           main] s.b.e.YamlPropertySourceLoader$Processor : Matched document with default matcher: {properties={sports={football=[Team1, Team2, Team3], basketball=[Team4, Team5, Team6]}, settings={football={period1=45, period2=90, players=11}, basketball={periods=4, players=5}}}}
2016-07-26 17:44:41.282 DEBUG 30015 --- [           main] s.b.e.YamlPropertySourceLoader$Processor : Loaded 1 document from YAML resource: class path resource [sports.yml]
{football=[Team1, Team2, Team3], basketball=[Team4, Team5, Team6]}
{football={period1=45, period2=90, players=11}, basketball={periods=4, players=5}}

在spring中操作配置属性

1 个答案: