有没有办法根据用户输入将值传递给观察者(这意味着传递的值不是一直都是固定的)?
from rx import Observable, Observer
def push_five_strings(observer,value):
observer.on_next(value)
#observer.on_next("Alpha")
observer.on_completed()
class PrintObserver(Observer):
def on_next(self, value):
print("Received {0}".format(value))
def on_completed(self):
print("Done!")
def on_error(self, error):
print("Error Occurred: {0}".format(error))
strings = [("Alpha", "Beta", "Gamma", "Delta", "Epsilon")]
for i in strings:
push_five_strings(strings) #e.g. getting the values to push in one string at a time from a list of strings
#push_five_strings("Gamma")
#push_five_strings("Alpha")
#push_five_strings("Beta")
#push_five_strings("Delta")
source = Observable.create(push_five_strings)
#source = Observable.from_(["Alpha", "Beta", "Gamma", "Delta", "Epsilon"])
#source = Observable.from_([value])
source.subscribe(PrintObserver())
我试图去了解RxPy,但网上几乎没有任何例子...
答案 0 :(得分:0)
puts d.run(['1','1','0','a'])
启动并输入结果:
from rx import Observable, Observer
import sys
class PrintObserver(Observer):
def on_next(self, value):
print("Received {0}".format(value))
def on_completed(self):
print("Done!")
def on_error(self, error):
print("Error Occurred: {0}".format(error))
Observable.from_(sys.stdin).subscribe(PrintObserver())
使用abc
Received abc
def
Received def
Done!
停止输入流。