Cassandra Hector:如何检索列族的所有行？

Question

我正在寻找一个代码示例来检索列族的所有行和所有列.就像是:

SELECT * FROM MyTable

我看到这可以使用RangeSlicesQuery完成,但你仍然需要提供一定的范围.我认为你也必须指定列名.有干净安全的方法吗？

使用Hector 1.0和Cassandra 1.0.

Answer 1

尝试这样的事情:

public class Dumper {
    private final Cluster cluster;
    private final Keyspace keyspace;

    public Dumper() {
        this.cluster = HFactory.getOrCreateCluster("Name", "hostname");
        this.keyspace = HFactory.createKeyspace("Keyspace", cluster, new QuorumAllConsistencyLevelPolicy());
    }

    public void run() {
        int row_count = 100;

        RangeSlicesQuery<UUID, String, Long> rangeSlicesQuery = HFactory
            .createRangeSlicesQuery(keyspace, UUIDSerializer.get(), StringSerializer.get(), LongSerializer.get())
            .setColumnFamily("Column Family")
            .setRange(null, null, false, 10)
            .setRowCount(row_count);

        UUID last_key = null;

        while (true) {
            rangeSlicesQuery.setKeys(last_key, null);
            System.out.println(" > " + last_key);

            QueryResult<OrderedRows<UUID, String, Long>> result = rangeSlicesQuery.execute();
            OrderedRows<UUID, String, Long> rows = result.get();
            Iterator<Row<UUID, String, Long>> rowsIterator = rows.iterator();

            // we'll skip this first one, since it is the same as the last one from previous time we executed
            if (last_key != null && rowsIterator != null) rowsIterator.next();   

            while (rowsIterator.hasNext()) {
              Row<UUID, String, Long> row = rowsIterator.next();
              last_key = row.getKey();

              if (row.getColumnSlice().getColumns().isEmpty()) {
                continue;
              }


              System.out.println(row);
            }

            if (rows.getCount() < row_count)
                break;
        }
    }

    public static void main(String[] args) {
        new Dumper().run();
    }
}

这将以100行的页面遍历列族.它每行只能获取10列(你也想要分页很长的行).

这适用于具有用于行键的uuids,用于列名称的字符串和用于值的long的列族.希望如何改变这一点应该是显而易见的.