scala - 从Scala Process流标准输出

Question

我正在使用scala Process启动python程序，并使用ProcessLogger从python程序捕获stdout。我看到python程序中的print语句仅在python程序完成后才打印。有没有办法在执行它们时流式处理python打印语句？

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不要使用-u在scala中缓冲您的输出，这将帮助您:

var cmd = "python -u python_test.py"