問題描述

先看我的代碼：（用于測試的）

People people1 = new People('join','21');People people2 = new People('jane','29');Observable.just(people1,people2).subscribeOn(AndroidSchedulers.mainThread()) //在主線程上把 UI 初始化.doOnNext(people -> { num = 0; show.setText(String.valueOf(num));}).repeat(4,Schedulers.newThread()) //在新線程上重復發送2次.observeOn(AndroidSchedulers.mainThread()) //在主線程上觀測.subscribe(peoples1 -> { String s = people1.toString(); num++; show.setText(String.valueOf(num)); Log.e('---',s);});

這里，just先發射第一個 People：people1，之后再發射people2.但是呢，Logcat如下：

06-29 20:25:20.251 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.251 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.281 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.281 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.287 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.287 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.302 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}06-29 20:25:20.302 32688-32688/demo.chestnut.com.test E/---: People{name=’join’, age=’21’}

這里打印了8次的 people1，按理由，不是應該打印4次 people1，打印4次的people2嗎？

問題解答

.subscribe(peoples1 -> { String s = people1.toString(); //問題出在這里的`people1` ，本該用`peoples1`的 num++; show.setText(String.valueOf(num)); Log.e('---',s);});