Python Gui 没有响应数据库

Question

这篇文章的目的是确定代码中的错误，我花了无数个小时试图更改代码以查看错误是什么。 我正在尝试通过 python GUI (tkinter) 将员工添加到数据库中。 目前它只插入“”信息，这是在我们的 def delete 中用打印语句检查的。 我在使用 def delete 时遇到了类似的问题，但是元组中的信息没有显示在列表框中。 有没有人有解决为什么会发生这些故障的方法？

def addEmp():
    top = Toplevel()
    top.title("Add Employee")
    top.geometry("470x385")
    app = Frame(top)
    app.grid()

    firstName = StringVar()
    lastName = StringVar()
    address = StringVar()
    payRate = StringVar()
    inUnion = StringVar()
    paymentMethod = StringVar()
    grp1 = StringVar()

    firstNameLabel = Label (app, text = "First Name:", font = 15)
    firstNameLabel.grid(row = 0, column = 0)
    firstNameEntry = Entry (app, text = "", textvariable = firstName)
    firstNameEntry.grid(row = 0, column = 1)
    surNameLabel = Label (app, text = "Surname:", font = 15)
    surNameLabel.grid(row = 1, column = 0)
    lastNameEntry= Entry (app, text = "", textvariable = lastName)
    lastNameEntry.grid(row = 1, column = 1)
    addressLabel = Label (app, text = "Address:", font = 15)
    addressLabel.grid(row = 2, column = 0)
    addressEntry = Entry (app, text = "", textvariable = address)
    addressEntry.grid(row = 2, column = 1)
    payRateLabel = Label (app, text = "Pay Rate Hourly:", font = 15)
    payRateLabel.grid(row = 3, column = 0)
    payRateEntry = Entry (app, text = "", textvariable = payRate)
    payRateEntry.grid(row = 3, column = 1)
    salaryLabel = Label (app, text = "Pay Rate Monthly:", font = 15)
    salaryLabel.grid(row = 4, column = 0)
    salaryEntry = Entry (app, text = "")
    salaryEntry.grid(row = 4, column = 1)
    payTypeLabel = Label (app, text = "Work Type:", font = 15)
    payTypeLabel.grid(row = 5, column = 0)

    radio1 = Radiobutton(app, text = "Full Time (Monthly Salary)", value = 'Monthly', variable = grp1)
    radio1.grid(row = 6, column = 1)
    radio2 = Radiobutton(app, text = "Part Time (Hourly Pay per Week)", value = 'Weekly', variable = grp1)
    radio2.grid(row = 7, column = 1)
    radio3 = Radiobutton(app, text = "Commission", value = 'Commission', variable = grp1)
    radio3.grid(row = 8, column = 1)
    paymentOptionsLabel = Label (app, text = "Payment Option:", font = 15)
    paymentOptionsLabel.grid(row = 9, column = 0)
    radio4 = Radiobutton(app, text = "Mail to current address", value = "Mail", variable = paymentMethod)
    radio4.grid(row = 10, column = 1)
    radio5 = Radiobutton(app, text = "Held by Paymaster", value = "Hold", variable = paymentMethod)
    radio5.grid(row = 11, column = 1)
    radio6 = Radiobutton(app, text = "Directly Deposit into bank account", value = "Bank", variable = paymentMethod)
    radio6.grid(row = 12, column = 1)
    unionLabel = Label (app, text = "Union:", font = 15)
    unionLabel.grid(row = 13, column = 0)
    radio1union = Radiobutton(app, text = "Yes", value = '1', variable = inUnion)
    radio1union.grid(row = 14, column = 1)
    radio2union = Radiobutton(app, text = "No", value = '0', variable = inUnion)
    radio2union.grid(row = 15, column = 1)


    submitButton = Button (app, command = addEmpSubmit, text = "Submit", font = 15)
    submitButton.grid(row = 20, column = 3)

def addEmpSubmit():
    c.execute("INSERT INTO employees (first_name, last_name, address, in_union, pay_type, payment_method) VALUES (?, ?, ?, ?, ?, ?)",
              (str(firstName), str(lastName), str(address), str(inUnion), str(payType), str(paymentMethod)))
    connection.commit()

    if payType == "Weekly":
        c.execute("UPDATE employees SET pay_rate=? WHERE employee_id=?", (payRate, c.lastrowid))
        connection.commit()
    elif payType == "Monthly":
        c.execute("UPDATE employees SET monthly_salary=? WHERE employee_id=?", (monthlySalary, c.lastrowid))
        connection.commit()
    elif payType == "Commission":
        c.execute("UPDATE employees SET monthly_salary=?, commission_rate=? WHERE employee_id=?",
                  (monthlySalary, commissionRate, c.lastrowid))
        connection.commit()


def deleteEmp():
    top = Toplevel()
    top.title("Delete Employee")
    top.geometry("470x385")
    deleteList = Listbox(app)
    deleteList.grid(row = 0, column = 0)
    listItem = ""

    c.execute("SELECT * FROM employees")
    allEmp = c.fetchall()
    for line in allEmp:
        for item in line:
            print(str(item))
            listItem += str(item)
        deleteList.insert(END, listItem)

    connection.commit()
    empIDLabel = Label (app, text = "Please select an Emp ID:", font = 15)
    empIDLabel.grid(row = 1, column = 0)
    empIDEntry = Entry (app, text = "")
    empIDEntry.grid(row = 2, column = 0)

    deleteButton = Button (app, text = "Delete", font = 15)
    deleteButton.grid(row = 2, column = 1)

Answer 1

1. 正确阅读有关ControlVariables / StringVar的文档。 您为INSERT INTO - 查询提供的值不是条目小部件的值。

>>> name=tk.StringVar()
>>> name.set("First Name")
>>> name
<Tkinter.StringVar instance at 0x01EB4CB0>
>>> str(name)
'PY_VAR0'
>>> name
<Tkinter.StringVar instance at 0x01EB4CB0>
>>> name.get()
'First Name'

2. 请确保您仅发布可以（至少）在您的系统上重现错误的代码示例。 对于这个原因，它是非常难以重现你有问题。 拆分您的代码并在较小的切片中对其进行调试。 自行检查不同的组件（例如，首先使用一些虚拟数据但使用变量、StringVars 等的 sql 部分）。

3. 您是否收到任何错误消息？ 如果是这样，请将它们添加到您的帖子中，以便我们可以查看您是否在某个时刻遇到困难。 类似于 - “我的代码不起作用 - 为什么？” 即使可能有正确的信息，也很难进行分析。