Question

我一直在尝试使用类库 appJar ，这是一个很好的Python基本用户界面。我有工作代码来执行我需要的功能，但是，当我围绕它运行UI时，它不会运行，并且UI对断点的反馈要少得多

相关代码在这里：

def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(fromDirectory):
    for file in files:
        with Archive.progressbar.progressbar(max_value=10) as progress:
            for i in range(1):
                filePath = os.path.join(root, file)
                ziph.write(filePath, relpath(filePath, ""))
                time.sleep(0.1)
                progress.update(i)
def beginBackup(btn):
    return app.questionBox("Ready?", "Click OK ")

上面设置了这些功能，下面应该运行它，但它没有

try:
    if (beginBackup == True):
        print(beginBackup)
        zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
        zipdir('tmp/', zipf)
        zipf.close()
        os.rename('Python.zip', "bak" + str(fileName) + ".zip")
        shutil.move("bak" + str(fileName) + ".zip", str(toDirectory) + "/bak" + str(fileName) + ".zip")
else:
    raise Exception("Something went wrong")
except Exception as e:
    app.warningBox("Error", "Something went wrong: {}".format(str(e)))

我觉得这是因为我将代码放在错误的位置，因为它感觉它永远不会实例化备份过程。

以下是完整代码：

import os
import zipfile
import shutil
import time
from os.path import relpath
from appJar import gui

# Global Variables
fromDirectory = ""
toDirectory = ""
fileName = time.time()
fileVersion = 1.2

# Setting up the mechanism
def backupsource(btn):
    fromDirectory = app.directoryBox(title="Source")

def backupdest(btn):
    toDirectory = app.directoryBox(title="Destination")

def zipdir(path, ziph):
    # ziph is zipfile handle
    for root, dirs, files in os.walk(fromDirectory):
        for file in files:
            with Archive.progressbar.progressbar(max_value=10) as progress:
                for i in range(1):
                    filePath = os.path.join(root, file)
                    ziph.write(filePath, relpath(filePath, ""))
                    time.sleep(0.1)
                    progress.update(i)
def beginBackup(btn):
    return app.questionBox("Ready?", "Click OK ")

# Open the GUI
app = gui()
app.showSplash("Simple zip v. 1.2", fill="grey", stripe="#ADDFAD", fg="white", font=44)

# Setup the visual styles of the app
app.setTitle("Simple Zip")
app.setIcon("img/logo.ico")
app.setGeometry(400, 300)
app.setResizable(canResize=True)

# Items inside of the GUI
app.addLabel("title", "Welcome to the simple backup utility")
app.setLabelBg("title", "gray")
app.addLabel("Backup", "Label goes here")

# Setup source buttons
app.addButton("Source", backupsource)
app.addButton("Destination", backupdest)

# Begin Backup section
app.addButton("Backup!", beginBackup)

# start the GUI
app.go()

try:
    if (beginBackup == True):
        print(beginBackup)
        zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
        zipdir('tmp/', zipf)
        zipf.close()
        os.rename('Python.zip', "bak" + str(fileName) + ".zip")
        shutil.move("bak" + str(fileName) + ".zip", str(toDirectory) + "/bak" + str(fileName) + ".zip")
else:
    raise Exception("Something went wrong")   

 except Exception as e:
        app.warningBox("Error", "Something went wrong: {}".format(str(e)))

Answer 1

第一个错误我发现的是os.renames，它并不存在。但我不能建议像这样发现错误。我用异常包装器重写了那个（并且短路是无用的rename命令，因为shutil.move可以处理重命名+在文件系统之间移动。

然后你做app.go()但没有任何东西通过这一行。它是GUI主循环。从现在起，appJar将调度事件。你无法控制主程序了。

所以一切都必须在这里完成（按下按钮时激活代码）：

def beginBackup(btn):
    return app.questionBox("Ready?", "Click OK ")

会弹出一个窗口而不再执行任何操作。它应该被完整的代码替换：

def beginBackup(btn):
    if not app.questionBox("Ready?", "Click OK"):
      app.warningBox("Error", "Cancelled")
    else:
      try:
        zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
        zipdir('tmp/', zipf)
        zipf.close()
        shutil.move('Python.zip', str(toDirectory) + "/bak" + str(fileName) + ".zip")
      except Exception as e:
        app.warningBox("Error", "Something went wrong: {}".format(str(e)))

所以现在如果在该回调中发生任何事情，你会得到一个很好的警告框，告诉你它究竟是什么（无法打开文件，语法错误等......）

如何通过UI包装器

1 个答案: