我试图将以下Scala 2.9隐式转换方法转换为2.10隐式类:
import java.sql.ResultSet
/**
* Implicitly convert a ResultSet to a Stream[ResultSet]. The Stream can then be
* traversed using the usual map, filter, etc.
*
* @param row the Result to convert
* @return a Stream wrapped around the ResultSet
*/
implicit def stream(row: ResultSet): Stream[ResultSet] = {
if (row.next) Stream.cons(row, stream(row))
else {
row.close()
Stream.empty
}
}
我的第一次尝试没有编译:
implicit class ResultSetStream(row: ResultSet) {
def stream: Stream[ResultSet] = {
if (row.next) Stream.cons(row, stream(row))
else {
row.close()
Stream.empty
}
}
}
我在stream(row)
上收到语法错误,因为stream
没有参数。
这样做的正确方法是什么?
答案 0 :(得分:5)
试试这个:
scala> import java.sql.ResultSet
import java.sql.ResultSet
scala> implicit class ResultSetStream(row: ResultSet) {
| def stream: Stream[ResultSet] = {
| if (row.next) Stream.cons(row, row.stream)
| else {
| row.close()
| Stream.empty
| }
| }
| }
defined class ResultSetStream
您将stream
定义为函数,因此stream(row)
无效。
您可以继承AnyVal
以创建Value Class并优化您的代码:
implicit class ResultSetStream(val row: ResultSet) extends AnyVal {
def stream: Stream[ResultSet] = {
if (row.next) Stream.cons(row, row.stream)
else {
row.close()
Stream.empty
}
}
}