Question

在python中，我们可以通过定义创建弹出窗口

def pop(msg):
    MessageBox = ctypes.windll.user32.MessageBoxW
    MessageBox(None, msg, 'Window title', 0)

然后

pop('some message')

将显示一个弹出窗口以通知用户。

但是如何使此弹出窗口位于所有其他窗口之上，以使用户绝对不会错过它？

Answer 1

尝试：

 at org.apache.hadoop.fs.RawLocalFileSystem.deprecatedGetFileStatus(RawLocalFileSystem.java:611)
        at org.apache.hadoop.fs.RawLocalFileSystem.getFileLinkStatusInternal(RawLocalFileSystem.java:824)
        at org.apache.hadoop.fs.RawLocalFileSystem.getFileStatus(RawLocalFileSystem.java:601)
        at org.apache.hadoop.fs.FilterFileSystem.getFileStatus(FilterFileSystem.java:428)
        at org.apache.hadoop.yarn.util.FSDownload.copy(FSDownload.java:253)
        at org.apache.hadoop.yarn.util.FSDownload.access$000(FSDownload.java:63)
        at org.apache.hadoop.yarn.util.FSDownload$2.run(FSDownload.java:361)
        at org.apache.hadoop.yarn.util.FSDownload$2.run(FSDownload.java:359)
        at java.security.AccessController.doPrivileged(Native Method)
        at javax.security.auth.Subject.doAs(Subject.java:421)
        at org.apache.hadoop.security.UserGroupInformation.doAs(UserGroupInformation.java:1762)
        at org.apache.hadoop.yarn.util.FSDownload.call(FSDownload.java:358)
        at org.apache.hadoop.yarn.util.FSDownload.call(FSDownload.java:62)
        at java.util.concurrent.FutureTask.run(FutureTask.java:262)
        at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:473)
        at java.util.concurrent.FutureTask.run(FutureTask.java:262)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1152)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:622)
        at java.lang.Thread.run(Thread.java:748)

Answer 2

使用tkinter可能是一个更好的主意-除了它是python的一部分之外，它也是跨平台的。

以下是弹出的示例：

from tkinter import messagebox, Toplevel, Label, Tk, Button


def pop(msg):  # This will be on top of the window which called this pop up
    messagebox.showinfo('message', msg)


def pop_always_on_top(msg):  # This will be on top of any other window
    msg_window = Toplevel()
    msg_window.title("Show Message")
    msg_window.attributes('-topmost', True)
    msg_label = Label(msg_window, text=msg)
    msg_label.pack(expand=1, fill='both')
    button = Button(msg_window, text="Ok", command=msg_window.destroy)
    button.pack()


if __name__ == '__main__':
    top = Tk()

    hello_btn = Button(top, text="Say Hello", command=lambda: pop('hello world!'))
    hello_btn.pack()
    hello_btn = Button(top, text="Say Hello and stay on top!", command=lambda: pop_always_on_top('hello world - on top!'))
    hello_btn.pack()
    top.mainloop()

该消息框自动位于其父窗口的顶部。

如果要使其位于系统上其他所有窗口的顶部，请设置窗口的“ -topmost”标志（如第二种方法所示）。

如何使python弹出窗口停留在顶部

2 个答案: