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python - 连接/处理脚本到 PySimpleGUI 按钮

转载 作者:行者123 更新时间:2023-12-01 07:31:02 25 4
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你们能帮我知道如何在我的 PySimpleGui 脚本中连接一个按钮,当按下/单击运行按钮时,该按钮将执行另一个 python 脚本。

现在,我一直在阅读有关 GUI 脚本中的 Subprocess 和 command = os.popen 的内容。

layout = [[ sg.Text('Click the button to launch Program')],
[sg.Button('Launch')]]

win1 = sg.Window('My new window').Layout(layout)


win2_activate = False

while True:
ev1, vals1 = win1.Read()
if ev1 is None or ev1 == 'Cancel':
break

if not win2_activate and ev1 == 'Launch':
win1.Hide()
win2_activate = True
layout2 = [[sg.Text('Report Auto')],
[sg.Input(do_not_clear=True)],
[sg.Text('', key='_OUTPUT_')],
[sg.Button('Run'), sg.Button('Cancel')]]

win2 = sg.Window('Window2').Layout(layout2)
while True:
ev2, vals2 = win2.Read()
if ev2 is None or ev2 =='Cancel':
win2_activate = False
win2.Close()
win1.UnHide()
break

在我的 pysimplegui 脚本中,我尚未包含子进程或任何库,因为我只是不知道在哪里执行它。非常欢迎任何帮助!

最佳答案

这是您问题的完整答案,一个 PySimpleGUI 程序。该程序允许您输入命令。然后按一个按钮。当按下按钮时,命令将“运行”,并且输出将显示在窗口中。

import subprocess
import sys
import PySimpleGUI as sg

def main():
layout = [ [sg.Text('Enter a command to execute (e.g. dir or ls)')],
[sg.Input(key='_IN_')], # input field where you'll type command
[sg.Output(size=(60,15))], # an output area where all print output will go
[sg.Button('Run'), sg.Button('Exit')] ] # a couple of buttons

window = sg.Window('Realtime Shell Command Output', layout)

while True: # Event Loop
event, values = window.Read()
if event in (None, 'Exit'): # checks if user wants to exit
break

if event == 'Run': # the two lines of code needed to get button and run command
runCommand(cmd=values['_IN_'], window=window)

window.Close()

# This function does the actual "running" of the command. Also watches for any output. If found output is printed
def runCommand(cmd, timeout=None, window=None):
p = subprocess.Popen(cmd, shell=True, stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
output = ''
for line in p.stdout:
line = line.decode(errors='replace' if (sys.version_info) < (3, 5) else 'backslashreplace').rstrip()
output += line
print(line)
window.Refresh() if window else None # yes, a 1-line if, so shoot me
retval = p.wait(timeout)
return (retval, output) # also return the output just for fun

if __name__ == '__main__':
main()

关于python - 连接/处理脚本到 PySimpleGUI 按钮,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57200315/

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