问题
I have a dataframe and I want to insert it into hbase. I follow this documenation .
This is how my dataframe look like:
--------------------
|id | name | address |
|--------------------|
|23 |marry |france |
|--------------------|
|87 |zied |italie |
--------------------
I create a hbase table using this code:
val tableName = "two"
val conf = HBaseConfiguration.create()
if(!admin.isTableAvailable(tableName)) {
print("-----------------------------------------------------------------------------------------------------------")
val tableDesc = new HTableDescriptor(tableName)
tableDesc.addFamily(new HColumnDescriptor("z1".getBytes()))
admin.createTable(tableDesc)
}else{
print("Table already exists!!--------------------------------------------------------------------------------------")
}
And now how may I insert this dataframe into hbase ?
In another example I succeed to insert into hbase using this code:
val myTable = new HTable(conf, tableName)
for (i <- 0 to 1000) {
var p = new Put(Bytes.toBytes(""+i))
p.add("z1".getBytes(), "name".getBytes(), Bytes.toBytes(""+(i*5)))
p.add("z1".getBytes(), "age".getBytes(), Bytes.toBytes("2017-04-20"))
p.add("z2".getBytes(), "job".getBytes(), Bytes.toBytes(""+i))
p.add("z2".getBytes(), "salary".getBytes(), Bytes.toBytes(""+i))
myTable.put(p)
}
myTable.flushCommits()
But now I am stuck, how to insert each record of my dataframe into my hbase table.
Thank you for your time and attention
回答1:
using answer for code formatting purposes Doc tells:
sc.parallelize(data).toDF.write.options(
Map(HBaseTableCatalog.tableCatalog -> catalog, HBaseTableCatalog.newTable -> "5"))
.format("org.apache.hadoop.hbase.spark ")
.save()
where sc.parallelize(data).toDF is your DataFrame. Doc example turns scala collection to dataframe using sc.parallelize(data).toDF
You already have your DataFrame, just try to call
yourDataFrame.write.options(
Map(HBaseTableCatalog.tableCatalog -> catalog, HBaseTableCatalog.newTable -> "5"))
.format("org.apache.hadoop.hbase.spark ")
.save()
And it should work. Doc is pretty clear...
UPD
Given a DataFrame with specified schema, above will create an HBase table with 5 regions and save the DataFrame inside. Note that if HBaseTableCatalog.newTable is not specified, the table has to be pre-created.
It's about data partitioning. Each HBase table can have 1...X regions. You should carefully pick number of regions. Low regions number is bad. High region numbers is also bad.
回答2:
An alternate is to look at rdd.saveAsNewAPIHadoopDataset, to insert the data into the hbase table.
def main(args: Array[String]): Unit = {
val spark = SparkSession.builder().appName("sparkToHive").enableHiveSupport().getOrCreate()
import spark.implicits._
val config = HBaseConfiguration.create()
config.set("hbase.zookeeper.quorum", "ip's")
config.set("hbase.zookeeper.property.clientPort","2181")
config.set(TableInputFormat.INPUT_TABLE, "tableName")
val newAPIJobConfiguration1 = Job.getInstance(config)
newAPIJobConfiguration1.getConfiguration().set(TableOutputFormat.OUTPUT_TABLE, "tableName")
newAPIJobConfiguration1.setOutputFormatClass(classOf[TableOutputFormat[ImmutableBytesWritable]])
val df: DataFrame = Seq(("foo", "1", "foo1"), ("bar", "2", "bar1")).toDF("key", "value1", "value2")
val hbasePuts= df.rdd.map((row: Row) => {
val put = new Put(Bytes.toBytes(row.getString(0)))
put.addColumn(Bytes.toBytes("cf1"), Bytes.toBytes("value1"), Bytes.toBytes(row.getString(1)))
put.addColumn(Bytes.toBytes("cf2"), Bytes.toBytes("value2"), Bytes.toBytes(row.getString(2)))
(new ImmutableBytesWritable(), put)
})
hbasePuts.saveAsNewAPIHadoopDataset(newAPIJobConfiguration1.getConfiguration())
}
Ref : https://sparkkb.wordpress.com/2015/05/04/save-javardd-to-hbase-using-saveasnewapihadoopdataset-spark-api-java-coding/
来源:https://stackoverflow.com/questions/44111988/insert-spark-dataframe-into-hbase