我有这个主线:
Gui.py
from gi.repository import Gtk, Gdk
import Process
import gobject
class gui():
def __init__(self):
self.window = Gtk.Window()
self.window.connect('delete-event', Gtk.main_quit)
self.box = Gtk.Box()
self.window.add(self.box)
self.label = Gtk.Label('idle')
self.box.pack_start(self.label, True, True, 0)
self.progressbar = Gtk.ProgressBar()
self.box.pack_start(self.progressbar, True, True, 0)
self.button = Gtk.Button(label='Start')
self.button.connect('clicked', self.on_button_clicked)
self.box.pack_start(self.button, True, True, 0)
self.window.show_all()
gobject.threads_init()
Gdk.threads_enter()
Gtk.main()
Gdk.threads_leave()
def working1():
self.label.set_text('working1')
t = Process.Heavy()
t.heavyworks1()
self.label.set_text('idle')
def on_button_clicked(self, widget):
Gdk.threads_enter()
working1()
Gdk.threads_leave()
if __name__ == '__main__':
gui = gui()
此代码将生成此gui: img http://i42.tinypic.com/33nvrx2.png
我有第二个模块可以做逻辑。
Process.py
import threading
class Heavy(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
def heavyworks1(self):
#doing heavy works1
#return result
def heavyworks2(self, *param):
#doing heavy works2
#return result
当我执行此操作时,操作正常,但gui变得冻结。怎么做得好?
修改
如user4815162342所说,我将我的代码更改为:
from gi.repository import Gtk, Gdk, GLib
import Process
import gobject
import threading
class gui():
def __init__(self):
self.window = Gtk.Window()
self.window.connect('delete-event', Gtk.main_quit)
self.box = Gtk.Box()
self.window.add(self.box)
self.label = Gtk.Label('idle')
self.box.pack_start(self.label, True, True, 0)
self.progressbar = Gtk.ProgressBar()
self.box.pack_start(self.progressbar, True, True, 0)
self.button = Gtk.Button(label='Start')
self.button.connect('clicked', self.on_button_clicked)
self.box.pack_start(self.button, True, True, 0)
self.window.show_all()
gobject.threads_init()
GLib.threads_init()
Gdk.threads_init()
Gdk.threads_enter()
Gtk.main()
Gdk.threads_leave()
def init_progress(self, func, arg):
self.label.set_text('working1')
self.worker = threading.Thread(target=func, args=[arg])
self.running = True
gobject.timeout_add(200, self.update_progress)
self.worker.start()
def update_progress(self):
if self.running:
self.progressbar.pulse()
return self.running
def working(self, num):
Process.heavyworks2(num)
gobject.idle_add(self.stop_progress)
def stop_progress(self):
self.running = False
self.worker.join()
self.progressbar.set_fraction(0)
self.label.set_text('idle')
def on_button_clicked(self, widget):
self.init_progress(self.working, 100000)
if __name__ == '__main__':
gui = gui()
使用该代码,程序有时会工作但有时会出现此错误。
1
**
Gtk:ERROR:/build/buildd/gtk+3.0-3.4.2/./gtk/gtktextview.c:3726:gtk_text_view_validate_onscreen: assertion failed: (priv->onscreen_validated)
Aborted (core dumped)
2
*** glibc detected *** python: free(): invalid next size (fast): 0x09c9f820 ***
3
Segmentation fault (core dumped)
答案 0 :(得分:4)
你实际上并没有启动线程,你只是实例化了一个可以用来启动它的对象。完整的解决方案需要仔细分离GUI线程和工作线程之间的职责。你想要做的是以下几点:
在单独的线程中进行繁重的计算,由GUI代码生成并加入。计算不应该生成自己的线程,也不需要知道线程(当然,除了线程安全之外)。
线程完成后,使用gobject.idle_add()
告诉GUI可以撤消进度指示器。 (gobject.idle_add
是唯一可以安全地从另一个线程调用的GTK函数。)
通过这样的设置,无论计算如何,GUI都保持完全响应并且进度条更新,并且GUI线程保证在计算完成时注意到。关于您当前代码的两点:
实例化threading.Thread
而不是继承它。这样您就不必费心实施run()
了。但是,在这两种情况下,您都必须调用thread.start()
来启动该主题。
请勿致电threads_enter()
和threads_leave()
,除非确实知道您在做什么。请记住,只要你从一个线程(你初始化GTK的同一个线程)中调用所有GTK函数,你就可以了。
以下是实现上述建议的概念验证代码:
def working1(self):
self.label.set_text('working1')
self.work_thread = threading.Thread(self.run_thread)
self.running = True
gobject.timeout_add(200, self.update_progress)
self.work_thread.start()
# the GUI thread now returns to the mainloop
# this will get periodically called in the GUI thread
def update_progress(self):
if self.running:
self.progressbar.pulse() # or set_fraction, etc.
return self.running
# this will get run in a separate thread
def run_thread(self):
Process.heavyworks1() # or however you're starting your calculation
gobject.idle_add(self.stop_progress)
# this will get run in the GUI thread when the worker thread is done
def stop_progress(self):
self.running = False
self.work_thread.join()
self.label.set_text('idle')
答案 1 :(得分:0)
正如您所建议的那样,您需要为此启动另一个线程。通常在python中进行线程非常简单,但使用GUI会变得棘手。