我正在尝试将我的应用程序连接到DynamoDB.我已按照亚马逊推荐的方式设置了一切.但我仍然一遍又一遍地得到同样的错误:
7-21 11:02:29.856 10027-10081/com.amazonaws.cognito.sync.demo E/AndroidRuntime? FATAL EXCEPTION: AsyncTask #1
Process: com.amazonaws.cognito.sync.demo, PID: 10027
java.lang.RuntimeException: An error occured while executing doInBackground()
at android.os.AsyncTask$3.done(AsyncTask.java:304)
at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:355)
at java.util.concurrent.FutureTask.setException(FutureTask.java:222)
at java.util.concurrent.FutureTask.run(FutureTask.java:242)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:818)
Caused by: com.amazonaws.services.dynamodbv2.model.ResourceNotFoundException: Requested resource not found (Service: AmazonDynamoDBv2; Status Code: 400; Error Code: ResourceNotFoundException; Request ID: GIONOKT7E3AMTC4PO19CPLON93VV4KQNSO5AEMVJF66Q9ASUAAJG)
at com.amazonaws.http.AmazonHttpClient.handleErrorResponse(AmazonHttpClient.java:710)
at com.amazonaws.http.AmazonHttpClient.executeHelper(AmazonHttpClient.java:385)
at com.amazonaws.http.AmazonHttpClient.execute(AmazonHttpClient.java:196)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.invoke(AmazonDynamoDBClient.java:2930)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.updateItem(AmazonDynamoDBClient.java:930)
at com.amazonaws.mobileconnectors.dynamodbv2.dynamodbmapper.DynamoDBMapper$SaveObjectHandler.doUpdateItem(DynamoDBMapper.java:1173)
at com.amazonaws.mobileconnectors.dynamodbv2.dynamodbmapper.DynamoDBMapper$2.executeLowLevelRequest(DynamoDBMapper.java:873)
at com.amazonaws.mobileconnectors.dynamodbv2.dynamodbmapper.DynamoDBMapper$SaveObjectHandler.execute(DynamoDBMapper.java:1056)
at com.amazonaws.mobileconnectors.dynamodbv2.dynamodbmapper.DynamoDBMapper.save(DynamoDBMapper.java:904)
at com.amazonaws.mobileconnectors.dynamodbv2.dynamodbmapper.DynamoDBMapper.save(DynamoDBMapper.java:688)
at com.amazonaws.cognito.sync.Utils.FriendsSyncManager.initalize_credentialprovider(FriendsSyncManager.java:43)
at com.amazonaws.cognito.sync.ValU.FriendListActivity$SyncFriends.doInBackground(FriendListActivity.java:168)
at com.amazonaws.cognito.sync.ValU.FriendListActivity$SyncFriends.doInBackground(FriendListActivity.java:160) …我目前正在开发一个 android 应用程序。这个应用程序包含一个 WebView 并显示一个我不拥有的网站。(Fe a WebStore)
该应用程序应该识别/检测用户何时单击本网站上的特定按钮（例如“购买产品”按钮）。如果用户点击这样的按钮，那么应用程序会弹出一个祝酒词。
我知道这对于每个网站都会有所不同，但是我通常如何对点击检测进行编程？我阅读了一些关于注入 JavaScript 等的文章，但是我真的不明白这应该如何工作。有没有办法说：如果用户点击 id=buy_button 的按钮，然后显示 toast？
谢谢你的帮助！