Python sys.stdin.read() 两次接受输入条件打印提示

Question

我写了一个函数，它接受一个字符（不点击enter），检查验证，并返回按下的键。但问题是，如果值不匹配，我正在打印的提示将打印两次。这是我的代码。

def accept_input():
    while True:
        print "Type Y to continue, ctrl-c to exit"
        ch = sys.stdin.read(1)
        if ch != "Y":
            pass
        else:
            return ch

当调用accept_input()时，如果有不匹配的字符则打印两次提示，如果输入为空白则打印一次。

python accept_input.py 
Type Y to continue, ctrl-c to exit
a
Type Y to continue, ctrl-c to exit
Type Y to continue, ctrl-c to exit
b
Type Y to continue, ctrl-c to exit
Type Y to continue, ctrl-c to exit
c
Type Y to continue, ctrl-c to exit
Type Y to continue, ctrl-c to exit

Type Y to continue, ctrl-c to exit

Type Y to continue, ctrl-c to exit
Y
accepted

为什么输入任何不匹配的键时打印两次，输入空白键时为什么只打印一次？

谢谢。

Answer 1

那是因为在a 之后你也按下了\n...所以2 个字符。你可以改为清除缓冲区。

def accept_input():
    import sys
    while True:
        print "Type Y to continue, ctrl-c to exit"
        ch = sys.stdin.read(1)
        sys.stdin.flush()   #<===========
        if ch != "Y":
            pass
        else:
            return ch