tar*_*rka 5 java iterator amazon-dynamodb
我正在对 DynamoDB 表执行扫描,然后需要将返回的项目中的相应属性添加到类型列表User(User具有单个构造函数User(String uuid))。该代码当前成功扫描数据库并返回List扫描结果。然而,由于某种原因,我的迭代似乎返回 null。
AmazonDynamoDBClient client = dynamoClient.getDynamoClient();
DynamoDBMapper mapper = new DynamoDBMapper(client);
try {
DynamoDBScanExpression scanExpression = new DynamoDBScanExpression();
Map<String, Condition> scanFilter = new HashMap<String, Condition>();
Condition scanCondition =
new Condition().withComparisonOperator(ComparisonOperator.NOT_NULL);
scanFilter.put("uuid", scanCondition);
scanExpression.setScanFilter(scanFilter);
List scanResults = mapper.scan(UserAccounts.class, scanExpression);
for (Iterator it = scanResults.iterator(); it.hasNext();) {
//User user = (User) it.next();
allUserSummary.add(new User(scanResults.get(1).toString()));
}
} catch (Exception e) {
// TODO
}
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我建议您开始通过For-Each Loop来使用现代且紧凑的列表迭代,这有助于避免使用旧迭代样式时的许多常见错误:
[...]
迭代器只是混乱。此外,这也是犯错误的机会。迭代器变量在每个循环中出现三次:这是出错的两次机会。for-each 结构消除了混乱和出错的机会。以下是使用 for-each 构造的示例的外观:
Run Code Online (Sandbox Code Playgroud)void cancelAll(Collection<TimerTask> c) { for (TimerTask t : c) t.cancel(); }
将其应用到您的用例中大约会产生以下结果:
List<UserAccounts> scanResults = mapper.scan(UserAccounts.class, scanExpression);
for (UserAccounts userAccounts : scanResults) {
allUserSummary.add(new User(userAccounts.toString()));
}
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如果这还不起作用,它也可能暗示实际的错误,只要您的代码假设 oftoString()类UserAccounts返回uuid,这可能是也可能不是。通常的方法是使用getKey()orgetUuidAttribute()方法和相应的注释@DynamoDBHashKeyor ,如DynamoDBMapper 类@DynamoDBAttribute的示例所示,例如:
@DynamoDBTable(tableName = "UserAccounts")
public class UserAccounts{
private String key; // or uuid right away
@DynamoDBHashKey
public String getKey() {
return key;
}
public void setKey(String key) {
this.key = key;
}
// ...
}
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对于您的示例,这显然会产生以下结果:
List<UserAccounts> scanResults = mapper.scan(UserAccounts.class, scanExpression);
for (UserAccounts userAccounts : scanResults) {
allUserSummary.add(new User(userAccounts.getKey()));
}
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