rxjava 七:zip操作符
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2024-02-28 12:11:10
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说明
zip:压缩工具生成的压缩文件个是
此处同理:将两个或多个事件压缩成一个事件处理
代码
Observable<Integer> observable1 = Observable.create(new ObservableOnSubscribe<Integer>() {
@Override
public void subscribe(ObservableEmitter<Integer> emitter) throws Exception {
emitter.onNext(1);
Log.i("zqq","1");
emitter.onNext(2);
Log.i("zqq","2");
emitter.onNext(3);
Log.i("zqq","3");
emitter.onNext(4);
Log.i("zqq","4");
emitter.onNext(5);
Log.i("zqq","5");
emitter.onComplete();
}
});
Observable<Integer> observable2 = Observable.create(new ObservableOnSubscribe<Integer>() {
@Override
public void subscribe(ObservableEmitter<Integer> emitter) throws Exception {
emitter.onNext(11);
Log.i("zqq","11");
emitter.onNext(22);
Log.i("zqq","22");
emitter.onNext(33);
Log.i("zqq","33");
emitter.onNext(44);
Log.i("zqq","44");
emitter.onComplete();
}
});
Observable.zip(observable1, observable2, new BiFunction<Integer, Integer, Integer>() {
@Override
public Integer apply(Integer integer, Integer integer2) throws Exception { //将两个observable1结合到一起,然后处理为新的结果
return integer+integer2;
}
})
.subscribe(new Consumer<Integer>() {
@Override
public void accept(Integer integer) throws Exception {
Log.i("zqq","accept>>"+integer);
}
});
结果:
可以看到:
1、observable1 和 observalbe2被组合到了一起
2、事件是严格按照顺序组合的
3、observable1 发送了5个onnext observalbe2 发送了4个,结果产生了4条,若事件无组合,忽略
再看:
observable1事件发送完成之后,observable2才开始发送,并且开始组合
原因:两个observable在通一线程,所以执行有先后顺序,解决方法,切换线程
如下,加延时,切线程
Observable<Integer> observable1 = Observable.create(new ObservableOnSubscribe<Integer>() {
@Override
public void subscribe(ObservableEmitter<Integer> emitter) throws Exception {
emitter.onNext(1);
Log.i("zqq","1");
Thread.sleep(1000);
emitter.onNext(2);
Log.i("zqq","2");
Thread.sleep(1000);
emitter.onNext(3);
Log.i("zqq","3");
Thread.sleep(1000);
emitter.onNext(4);
Log.i("zqq","4");
Thread.sleep(1000);
emitter.onNext(5);
Log.i("zqq","5");
Thread.sleep(1000);
emitter.onComplete();
}
}).subscribeOn(Schedulers.io());
Observable<Integer> observable2 = Observable.create(new ObservableOnSubscribe<Integer>() {
@Override
public void subscribe(ObservableEmitter<Integer> emitter) throws Exception {
emitter.onNext(11);
Log.i("zqq","11");
Thread.sleep(1000);
emitter.onNext(22);
Log.i("zqq","22");
Thread.sleep(1000);
emitter.onNext(33);
Log.i("zqq","33");
Thread.sleep(1000);
emitter.onNext(44);
Log.i("zqq","44");
Thread.sleep(1000);
emitter.onComplete();
}
}).subscribeOn(Schedulers.io());
Observable.zip(observable1, observable2, new BiFunction<Integer, Integer, Integer>() {
@Override
public Integer apply(Integer integer, Integer integer2) throws Exception { //将两个observable1结合到一起,然后处理为新的结果
return integer+integer2;
}
})
.subscribeOn(Schedulers.io())
.subscribe(new Consumer<Integer>() {
@Override
public void accept(Integer integer) throws Exception {
Log.i("zqq","accept>>"+integer);
}
});
结果:
实际开发中,经常遇到,一个列表,一个页面,多个数据组合的问题,可以用此种方式来处理
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