PyQt: best way to do the trick "start at boot" for my program in Windows
I'm usi开发者_高级运维ng PyQt to develop an application that in Windows, if set in preferences, should be able to start at boot.
I'm releasing this software with PyInstaller as a single executable file; i don't have a proper "installer".
Which is the best way to achieve this? ( = starting at boot)
A possible solution is to add a link in the startup folder, but i have to do it from the software: it's possible? Other ways?
There is an universal path to the Startup folder? Can i have some rights' problem?
try this code (it works for me with py2exe):
import sys
from PyQt4.QtCore import QSettings
from PyQt4.QtGui import (QApplication, QWidget, QCheckBox, QPushButton,
QVBoxLayout)
RUN_PATH = "HKEY_CURRENT_USER\\Software\\Microsoft\\Windows\\CurrentVersion\\Run"
class MainWidget(QWidget):
def __init__(self,parent=None):
super(MainWidget, self).__init__(parent)
self.settings = QSettings(RUN_PATH, QSettings.NativeFormat)
self.setupUi()
# Check if value exists in registry
self.checkbox.setChecked(self.settings.contains("MainWidget"))
def setupUi(self):
self.checkbox = QCheckBox("Boot at Startup", self)
button = QPushButton("Close", self)
button.clicked.connect(self.close)
layout = QVBoxLayout(self)
layout.addWidget(self.checkbox)
layout.addWidget(button)
def closeEvent(self, event):
if self.checkbox.isChecked():
self.settings.setValue("MainWidget",sys.argv[0]);
else:
self.settings.remove("MainWidget");
event.accept()
if __name__ == '__main__':
app = QApplication(sys.argv)
w = MainWidget()
w.show()
app.exec_()
You may add registry key under [HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Run], with any name and value "path_to_your_exec". this will require local administrator right, but will work for all users. The same key but starting with [HKEY_CURRENT_USER] will not require administrator privileges, but will work only for current user. That registry path is the same for win2k..win7
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