SparkSQL创建RDD：<4>动态创建Schema将非json格式的RDD转换成DataFrame【Java，Scala纯代码】

Java版本 

SparkConf conf = new SparkConf();
conf.setMaster("local").setAppName("rddStruct");
JavaSparkContext sc = new JavaSparkContext(conf);
SQLContext sqlContext = new SQLContext(sc);
JavaRDD<String> lineRDD = sc.textFile("./sparksql/person.txt");
/**
 * 转换成Row类型的RDD
 */
JavaRDD<Row> rowRDD = lineRDD.map(new Function<String, Row>() {

	/**
	 * 
	 */
	private static final long serialVersionUID = 1L;

	@Override
	public Row call(String s) throws Exception {
          return RowFactory.create(
                String.valueOf(s.split(",")[0]),
                String.valueOf(s.split(",")[1]),
                Integer.valueOf(s.split(",")[2])
	);
	}
});
/**
 * 动态构建DataFrame中的元数据，一般来说这里的字段可以来源自字符串，也可以来源于外部数据库
 */
List<StructField> asList =Arrays.asList(
	DataTypes.createStructField("id", DataTypes.StringType, true),
	DataTypes.createStructField("name", DataTypes.StringType, true),
	DataTypes.createStructField("age", DataTypes.IntegerType, true)
);

StructType schema = DataTypes.createStructType(asList);
DataFrame df = sqlContext.createDataFrame(rowRDD, schema);

df.show();
sc.stop();

Scala版代码：

val conf = new SparkConf()
conf.setMaster("local").setAppName("rddStruct")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)
val lineRDD = sc.textFile("./sparksql/person.txt")
val rowRDD = lineRDD.map { x => {
  val split = x.split(",")
  RowFactory.create(split(0),split(1),Integer.valueOf(split(2)))
} }

val schema = StructType(List(
  StructField("id",StringType,true),
  StructField("name",StringType,true),
  StructField("age",IntegerType,true)
))

val df = sqlContext.createDataFrame(rowRDD, schema)
df.show()
df.printSchema()
sc.stop()

鼓励一下我呗，谢谢你。