Android + RxJava + For Loop + 未执行所有请求

huangapple
go评论118阅读模式
英文:

Android + RxJava + For Loop + Not executing all the requests

问题

调用方法，

  1. for (String name : controllerToPartitionModels.keySet()) {
  2.     List<PartitionModel> partitionsList = controllerToPartitionModels.get(name);
  3.     refreshPartition(partitionsList, false);
  4. }

方法

  1. private void refreshPartition(List<PartitionModel> partitionModels, boolean isSyncAll) {
  2.     ITModule.getITService()
  3.             .refreshPartitionStatus(new ArrayList<>(partitionModels), isSyncAll)
  4.             .subscribeOn(Schedulers.io())
  5.             .observeOn(Schedulers.io())
  6.             .subscribe(new Action() {
  7.                 @Override
  8.                 public void run() throws Exception {
  9.                     Logger.get().d(ATTActionManager.this, "Refreshing request sent successfully for list of size : " + partitionModels.size());
  10.                 }
  11.             }, (@NonNull Throwable throwable) -> {
  12.                 Logger.get().d(ATTActionManager.this, "Error on Refresh request");
  13.             });
  14. }

问题
如果有2个请求需要发送，有时我只会看到一个请求被发送。也就是说，即使for循环执行了两次以进行2次请求（HTTP），我只会看到一个请求被发送到服务器。
我在这里做错了什么？
RxJava 版本使用：2.2.19

英文:

Caller Of the method,

  1.  for (String name : controllerToPartitionModels.keySet())
  2.     {
  3.         List&lt;PartitionModel&gt; partitionsList = controllerToPartitionModels.get(name);
  4.         refreshPartition(partitionsList,false);
  5.     }

Method

  1.  private void refreshPartition(List&lt;PartitionModel&gt; partitionModels, boolean isSyncAll) {
  2.     ITModule.getITService()
  3.             .refreshPartitionStatus(new ArrayList&lt;&gt;(partitionModels), isSyncAll)
  4.             .subscribeOn(Schedulers.io())
  5.             .observeOn(Schedulers.io())
  6.             .subscribe(new Action() {
  7.                 @Override
  8.                 public void run() throws Exception {
  9.                     Logger.get().d(ATTActionManager.this, &quot;Refreshing request sent successfully for list of size : &quot; + partitionModels.size());
  10.                 }
  11.             }, (@NonNull Throwable throwable) -&gt; {
  12.                 Logger.get().d(ATTActionManager.this, &quot;Error on Refresh request&quot;);
  13.             });
  14. }

Problem
If there are 2 requests that has to be sent, I sometime see only one request being sent. Meaning, even though for loop is executing twice for 2 request(HTTP), I see only one request is being sent to the server.
What is that i am doing wrong here?
Rxjava version in use : 2.2.19

答案1

得分: 2

你可以使用 flatMapIterable 合并上述的两种方法来解决你的问题。

合并后的解决方案：

  1. private void refreshPartition(Map<String, ?> controllerToPartitionModels) {
  3.     Observable.just(controllerToPartitionModels)
  4.             .map(controllerToPartitionModels -> controllerToPartitionModels.keySet())
  5.             .flatMapIterable((Function<Set<String>, Iterable<String>>) name -> name)
  6.             .map(name -> {
  7.                 boolean isSyncAll = false; // 你可以根据需求进行自定义
  8.                 return new Pair<List<PartitionModel>, Boolean>(controllerToPartitionModels.get(name), isSyncAll);
  9.             })
  10.             .flatMap((Function<Pair<List<PartitionModel>, Boolean>, ObservableSource<?>>) pair -> {
  11.                         boolean isSyncAll = pair.second;
  12.                         List<PartitionModel> partitionModels = pair.first;
  13.                         return ITModule.getITService()
  14.                                 .refreshPartitionStatus(new ArrayList<>(partitionModels), isSyncAll);
  15.                     }
  16.             )
  17.             .subscribeOn(Schedulers.io())
  18.             .observeOn(Schedulers.io())
  19.             .subscribe(() -> {
  20.                 Logger.get().d(ATTActionManager.this, "刷新请求已成功发送，列表大小为: " + partitionModels.size());
  21.             }, (@NonNull Throwable throwable) -> {
  22.                 Logger.get().d(ATTActionManager.this, "刷新请求出错");
  23.             });
  25. }

*请将 ? 替换为有效的对象类型。

英文:

You can merge the above 2 methods to solve your problem by using flatMapIterable.

Merged Solution:

  1. private void refreshPartition(Map&lt;String, ?&gt; controllerToPartitionModels) {
  3.     Observable.just(controllerToPartitionModels)
  4.             .map(controllerToPartitionModels -&gt; controllerToPartitionModels.keySet())
  5.             .flatMapIterable((Function&lt;Set&lt;String&gt;, Iterable&lt;String&gt;&gt;) name -&gt; name)
  6.             .map(name -&gt; {
  7.                 boolean isSyncAll = false; // You can customise as per requirement
  8.                 return new Pair&lt;List&lt;PartitionModel&gt;, Boolean&gt;(controllerToPartitionModels.get(name), isSyncAll)
  9.             })
  10.             .flatMap((Function&lt;Pair&lt;List&lt;PartitionModel&gt;, Boolean&gt;, ObservableSource&lt;?&gt;&gt;) pair -&gt; {
  11.                         boolean isSyncAll = pair.first;
  12.                         List&lt;PartitionModel&gt; partitionModels = pair.second;
  13.                         return ITModule.getITService()
  14.                                 .refreshPartitionStatus(new ArrayList&lt;&gt;(partitionModels), isSyncAll)
  15.                     }
  16.             )
  17.             .subscribeOn(Schedulers.io())
  18.             .observeOn(Schedulers.io())
  19.             .subscribe(new Action() {
  20.                 @Override
  21.                 public void run() throws Exception {
  22.                     Logger.get().d(ATTActionManager.this, &quot;Refreshing request sent successfully for list of size : &quot; + partitionModels.size());
  23.                 }
  24.             }, (@NonNull Throwable throwable) -&gt; {
  25.                 Logger.get().d(ATTActionManager.this, &quot;Error on Refresh request&quot;);
  26.             });
  28. }

*Kindly replace ? with the valid object type.

huangapple
  • 本文由 发表于 2020年8月3日 21:10:59
  • 转载请务必保留本文链接：https://go.coder-hub.com/63230087.html
  • android
  • java
  • performance
  • rx-java
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定