英文:
Reading application.properties in static method
问题
我在`SpringApplicationBuilder`中静态地放置了一些属性的代码如下：
@SpringBootApplication(scanBasePackages = { "com" })
public class Application extends SpringBootServletInitializer {
public static void main(String[] args) {
SpringApplicationBuilder parentBuilder = new SpringApplicationBuilder(Application.class);
parentBuilder.child(RestConfiguration.class, SwaggerConfig.class)
.properties("server.port:9093")).web(WebApplicationType.SERVLET).run(args);
}
}
我想将这些属性移动到`application.properties`文件中：
application.port.query=9093
我使用了`@Value`从应用程序文件中读取，但是我得到了null。在静态方法中读取数据还有其他方法吗？
英文:
I have the below code when i put some properties statically in SpringApplicationBuilder
@SpringBootApplication(scanBasePackages = { "com" })
public class Application extends SpringBootServletInitializer {
public static void main(String[] args) {
SpringApplicationBuilder parentBuilder = new SpringApplicationBuilder(Application.class);
parentBuilder.child(RestConfiguration.class, SwaggerConfig.class)
.properties("server.port:9093")).web(WebApplicationType.SERVLET).run(args);
}
}
I want to move the properties to the file application.properties
application.port.query=9093
I used @value to read from the application file, but i get null. Is there another way to read data in a static method?
答案1
得分: 1
尝试类似这样的代码：
@Configuration
@ConfigurationProperties("application")
class A {
public static int queryPort;
@Value("${port.query:9093}")
public void setQueryPort(final int portQuery){
A.queryPort = portQuery;
}
}
public class Application extends SpringBootServletInitializer {
public static void main(String[] args) {
SpringApplicationBuilder parentBuilder = new SpringApplicationBuilder(Application.class);
parentBuilder.child(RestConfiguration.class, SwaggerConfig.class)
.properties(A.queryPort)).web(WebApplicationType.SERVLET).run(args);
}
}
英文:
try something like this
@Configuration
@ConfigurationProperties("application")
class A {
public static int queryPort;
@Value("${port.query:9093}")
public void setQueryPort(final int portQuery){
A.queryPort = portQuery;
}
}
public class Application extends SpringBootServletInitializer {
public static void main(String[] args) {
SpringApplicationBuilder parentBuilder = new SpringApplicationBuilder(Application.class);
parentBuilder.child(RestConfiguration.class, SwaggerConfig.class)
.properties(A.queryPort)).web(WebApplicationType.SERVLET).run(args);
}
}
答案2
得分: 0
以下是翻译好的部分：
你正在尝试实现的首选方法是使用 Spring Bean 本身，它是你所有配置的父级 org.springframework.core.env.Environment。
@SpringBootApplication(scanBasePackages = {"com"})
public class Application {
public static void main(String[] args) {
ConfigurableApplicationContext applicationContext =
SpringApplication.run(Application.class, args);
Environment environment = applicationContext.getBean(Environment.class);
String propertyValue = environment.getProperty("any.property.from.configuration");
}
}
在你的用例中，只需获取 Environment 的 Bean，然后从 Environment 对象中获取任何属性值。
英文:
The preferable way to what you are trying to achieve is to use the spring bean itself which is the parent of your all configurations org.springframework.core.env.Environment
@SpringBootApplication(scanBasePackages = {"com"})
public class Application {
public static void main(String[] args) {
ConfigurableApplicationContext applicationContext =
SpringApplication.run(Application.class, args);
Environment environment = applicationContext.getBean(Environment.class);
String propertyValue = environment.getProperty("any.property.from.configuration")
}
}
In your use-case, just get the bean of Environment and from Environment object get any property value.
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