scala流程的流标准输出

Question

我使用scala进程来启动python程序，并使用ProcessLogger从python程序中捕获stdout。我看到python程序中的print语句只有在python程序完成后才会打印出来。是否有方法在执行python print语句时对其进行流处理？

import scala.sys.process.{ProcessLogger, _}

object TestProcessStdOut {
  def main(args: Array[String]) {
    var cmd = "python python_test.py";
    val process = Process(cmd).run(new ProcessLogger {
      override def out(s: => String) = println(s)

      override def buffer[T](f: => T) = ???

      override def err(s: => String) = ???
    })
  }
}

python_test.py

import time

print("print data 1")
time.sleep(2)
print("print data 2")
time.sleep(2)
print("print data 3")
time.sleep(2)

Answer 1

告诉python不要用-u缓冲scala中的输出将有助于您：

var cmd = "python -u python_test.py"