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如果我有以下蝟命令:Hystrix是否可以根據方法參數打開電路?
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 + "!";
}
}
被稱爲:
// no timeout for "Quick"
String s1 = new TimeoutDependingOnParam("Quick").execute();
// timeout for "Looong"
String s2 = new TimeoutDependingOnParam("Looong").execute();
如果蝟斷開電路,因爲與「Looong」超時通話,是否意味着通話用「快速」將打開到?
謝謝@ahus,所以覺得我可以做somethink像'publicTimeoutDependingOnParam(字符串名稱){超(HystrixCommandGroupKey.Factory.asKey( 「TDPGroup」))andCommandKey(HystrixCommandKey.Factory.asKey(」。 TDP「+ name))); this.name = name;}' – Pleymor
確切地說,應該可以工作。 – ahus1