我是否需要 DB 保留用户从 Tkinter GUI 输入的值？

Question

我正在使用一种 SEO 工具并尝试创建一个 GUI 应用程序，用户将在其中进入网站 link。然后，我的 list_of_url() 函数将执行以查找所有存在的 link。

```python
def ui_for_website():
    app = Tk()
    app.title("Scrapper")
    app.geometry("300x300")
    
    ui_text=StringVar(app,name="str")
    ui_text_label=Label(app,text="Enter your website: ",font=('bold',12),pady=20)
    ui_text_label.grid(row=0,column=0)
    ui_text_entry=Entry(app,textvariable=ui_text)
    ui_text_entry.grid(row=0,column=1)
    #onclick=lambda : ui_text_entry.get()
    ui_text_button=Button(app,text="Submit",command=lambda : onClick(ui_text_entry,app))

    ui_text_button.grid(row=2,column=1)
    
    ui_text=ui_text_entry.get()
    print(ui_text)
    app.mainloop()
```

这是我要求用户输入的 Tkinter UI 代码。

```python
def onClick(entry,app):
    #global website_name
    website_name=entry.get()
    #print(website_name)
    #app.quit()
    return website_name
``

这是我的 onClick 方法，一旦用户单击“提交”按钮，它就会执行。我正在尝试 return 值并将其存储在 website_name 变量中。

```python
def list_of_url():
    
    url=ui_for_website()
    print(url)
    urls=session.get(url).html.absolute_links
    
    tree = sitemap_tree_for_homepage(url)
    #print(len(tree.all_pages()))
    #print(len(tree.pages))
    url_list=[]
    for page in tree.all_pages():
        if ('admin' not in page.url) and (page.url not in url_list):
            url_list.append(page.url)

    print(len(url_list))
    return url_list
```

这个website_name，我在这个list_of_urls() 方法中访问它。但是有一次，tkinter 被关闭了。此变量的值变为空，我无法获取该值。

我得到的错误是

Traceback (most recent call last):
  File "Indexing.py", line 89, in <module>
    list_of_url()
  File "Indexing.py", line 42, in list_of_url
    urls=session.get(url).html.absolute_links
  File "C:\Users\<username>\Anaconda3\lib\site-packages\requests\sessions.py", line 546, in get
    return self.request('GET', url, **kwargs)
  File "C:\Users\<username>\Anaconda3\lib\site-packages\requests\sessions.py", line 519, in request
    prep = self.prepare_request(req)
  File "C:\Users\<username>\Anaconda3\lib\site-packages\requests\sessions.py", line 462, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "C:\Users\<username>\Anaconda3\lib\site-packages\requests\models.py", line 313, in prepare
    self.prepare_url(url, params)
  File "C:\Users\<username>\Anaconda3\lib\site-packages\requests\models.py", line 387, in prepare_url
    raise MissingSchema(error)
requests.exceptions.MissingSchema: Invalid URL 'None': No schema supplied. Perhaps you meant http://None?

因此，这让我想到是否需要数据库来保留用户输入值。这样即使GUI被销毁或关闭。我会保留用户输入的值。

这是我的完整代码：

```python
from requests_html import HTMLSession
from tkinter import *

from usp.tree import sitemap_tree_for_homepage

session=HTMLSession()
#website_name=''

def onClick(entry,app):
    #global website_name
    website_name=entry.get()
    print(website_name)
    #app.quit()
    return website_name

def ui_for_website():
    app = Tk()
    app.title("Scrapper")
    app.geometry("300x300")
    
    ui_text=StringVar(app,name="str")
    ui_text_label=Label(app,text="Enter your website: ",font=('bold',12),pady=20)
    ui_text_label.grid(row=0,column=0)
    ui_text_entry=Entry(app,textvariable=ui_text)
    ui_text_entry.grid(row=0,column=1)
    #onclick=lambda : ui_text_entry.get()
    ui_text_button=Button(app,text="Submit",command=lambda : onClick(ui_text_entry,app))

    ui_text_button.grid(row=2,column=1)
    
    ui_text=ui_text_entry.get()
    print(ui_text)
    app.mainloop()

def list_of_url():
    
    url=ui_for_website()
    print(url)
    urls=session.get(url).html.absolute_links
    
    tree = sitemap_tree_for_homepage(url)
    #print(len(tree.all_pages()))
    #print(len(tree.pages))
    url_list=[]
    for page in tree.all_pages():
        if ('admin' not in page.url) and (page.url not in url_list):
            url_list.append(page.url)

    print(len(url_list))
    return url_list
```

Answer 1

Tkinter 不需要连接到数据库。错误消息是关于 request 到 http。更具体地说，它来自 requests html 库。不涉及数据库。

我是否需要 DB 保留用户从 Tkinter GUI 输入的值？

Do i require DB to retain user-entered value from Tkinter GUI?

python

tkinter

python-3.x

tkinter-entry