选择在运行时弹出时要注入的实现

Question

我有以下课程:

public interface MyInterface{}

public class MyImpl1 implements MyInterface{}

public class MyImpl2 implements MyInterface{}

public class Runner {
        @Autowired private MyInterface myInterface;
}

我想做的是决定,虽然应用程序已经运行(即不在启动时)应该注入实现Runner.

所以理想情况是这样的:

ApplicationContext appContext = ...
Integer request = ...

Runner runner = null;
if (request == 1) {
        //here the property 'myInterface' of 'Runner' would be injected with MyImpl1
        runner = appContext.getBean(Runner.class) 
}
else if (request == 2) {
        //here the property 'myInterface' of 'Runner' would be injected with MyImpl2
        runner = appContext.getBean(Runner.class)
}
runner.start();

完成此任务的最佳方法是什么？

Answer 1

用@Component("implForRq1")和声明实现@Component("implForRq2")

然后注入它们并使用:

class Runner {

    @Autowired @Qualifier("implForRq1")
    private MyInterface runnerOfRq1;

    @Autowired @Qualifier("implForRq2")
    private MyInterface runnerOfRq2;

    void run(int rq) {
        switch (rq) {
            case 1: runnerOfRq1.run();
            case 2: runnerOfRq2.run();
            ...

        }
    }

}

...

@Autowired
Runner runner;

void run(int rq) {
    runner.run(rq);
}