Hystrix 是否能够根据方法参数打开电路？

Question

如果我有以下 Hystrix 命令：

public class TimeoutDependingOnParam extends HystrixCommand<String> {

    private final String name;

    public TimeoutDependingOnParam (String name) {
        super(HystrixCommandGroupKey.Factory.asKey("ExampleGroup"));
        this.name = name;
    }

    @Override
    protected String run() {

        if (name.equals("Looong")) {
           waitABillionYears();
        }

        return "Hello " + name + "!";
    }
}

调用者：

// no timeout for "Quick"
String s1 = new TimeoutDependingOnParam("Quick").execute();

// timeout for "Looong"
String s2 = new TimeoutDependingOnParam("Looong").execute();

如果 Hystrix 因“Looong”调用超时而打开电路，这是否意味着“Quick”调用将被打开？

Answer 1

只要两者都具有与您的示例中相同的命令键，基本上可以。但要让断路器分闸还有更多条件as stated in the documentation about the Circuit Breaker。

您可以实现两个不同的命令，也可以根据参数在构造函数中设置 CommandKey。 This is an extract from the documentation:

public CommandHelloWorld(String name) {
    super(Setter.withGroupKey(HystrixCommandGroupKey.Factory.asKey("ExampleGroup"))
            .andCommandKey(HystrixCommandKey.Factory.asKey("HelloWorld")));
    this.name = name;
}