用scala對accumulo表進行迭代

Question

我有一個accumulo命名記錄的表，每個row_id有幾個族和限定符，它看起來就像在accumulo shell中一樣。

michaelp@accumulo records> scan 
2016-10-17 16:27:55,359 [Shell.audit] INFO : michaelp@accumulo records> scan 
E001 department:sales [] 0 
E001 hire_date:20160101 [] 0 
E001 name:bob [] 0 
E001 name:jerry [] 0 
E002 department:marketing [] 0 
E002 hire_date:20160202 [] 0 
E002 name:sarah [] 0 
E003 department:engineering [] 0 
E003 hire_date:20160303 [] 0 
E003 name:joe [] 0 

我希望能夠用scala連接器掃描這幾行。所需的進口後，我的代碼如下所示：

var opts = new ClientOnRequiredTable() 
var bsOpts = new BatchScannerOpts() 
opts.parseArgs("test", Array("-t", "records","-u", "michaelp", "-p", "****", "-z", "zookeeper:2181", "-i", "accumulo"), bsOpts) 
var connector = opts.getConnector() 
var batchReader = connector.createBatchScanner("records", opts.auths, bsOpts.scanThreads) 
batchReader.setTimeout(bsOpts.scanTimeout, TimeUnit.MILLISECONDS) 
var x = new Range() 
var y = new LinkedList[Range] 
y.add(x) 
batchReader.setRanges(y) 

我通過在一個空的範圍內讓每一個行的表。問題是當我嘗試遍歷結果。它堅持在第一行。

scala> while (batchReader.iterator.hasNext()) {println(batchReader.iterator.next.getKey().toString())} 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
E001 department:sales [] 1476720996135 false 
... 

那麼爲什麼迭代器不能一起移動呢？

Answer 1

因爲當您每次調用batchReader.iterator時都會創建新的迭代器。而不是像下面這樣做

val iterator = batchReader.iterator 

while(iterator.hasNext) { 
println(iterator.next.getKey().toString()) 
}