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尝试使用失败。destroy()从GUI擦除数据结果

tkinter python

EdMed · 技术社区 · 3 年前

我希望当用户选择一个按钮时,它会破坏所有显示的表结果。目前,表格结果仍然没有错误。根据下面的代码片段,有什么突出的地方吗?

global searched_label
searched_label.destroy()

search = tracking_id_entry.get()
sql = "SELECT * FROM table_desc WHERE tracking_id = %s"
name = (search,)
my_cursor.execute(sql, name)
result = my_cursor.fetchall()

if not result:
    result = "Tracking ID Not Found..."
    searched_label = Label(screen, text=result)
    searched_label.grid(row=7, columnspan=2, sticky=EW)

else:
    for index, x in enumerate(result):
        num = 0
        index += 7
        for y in x:
            searched_label = Label(screen, text=y)
            searched_label.grid(row=index, column=num)
            num += 1

1 回复 | 直到 3 年前

acw1668 3 年前

因为您使用了相同的变量 searched_label 对于所有标签,只有最后一个标签将被该行销毁 searched_label.destroy()

您可以使用列表存储创建的标签,然后通过此列表销毁所有标签:

labels = []  # list to store the labels

# assume the posted code is inside a function
def show_tracking():
    # destroy existing labels
    for lbl in labels:
        lbl.destroy()
    labels.clear() # clear the list

    # populate new labels
    search = tracking_id_entry.get()
    sql = "SELECT * FROM table_desc WHERE tracking_id = %s"
    name = (search,)
    my_cursor.execute(sql, name)
    result = my_cursor.fetchall()

    if not result:
        result = "Tracking ID Not Found..."
        searched_label = Label(screen, text=result)
        searched_label.grid(row=7, columnspan=2, sticky=EW)
        labels.append(searched_label) # add the label to the list

    else:
        for index, x in enumerate(result):
            num = 0
            index += 7
            for y in x:
                searched_label = Label(screen, text=y)
                searched_label.grid(row=index, column=num)
                labels.append(searched_label) # add the label to the list
                num += 1

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