sqlite python數據庫錯誤 - 沒有這樣的表

Question

我用一個名爲product的表做了一個數據庫。當我運行下面的代碼時，出現錯誤 - sqlite3.OperationalError：no such table：product。我已經使用數據庫瀏覽器檢查並確定表存在。有任何想法嗎？代碼和文件都在同一個文件夾中。由於

from tkinter import * 
from tkinter import ttk 
import sqlite3 
import os.path 

class Product: 
    db_name = 'database.db' 

    def __init__(self, wind): 
     self.wind = wind 
     self.wind.title('IT Products') 

     frame = LabelFrame (self.wind, text = 'Add new record') 
     frame.grid (row = 0, column = 1) 

     Label (frame, text = 'Name: ').grid (row = 1, column = 1) 
     self.name = Entry (frame) 
     self.name.grid(row = 1, column = 2) 

     Label (frame, text = 'Price: ').grid (row = 2, column = 1) 
     self.price = Entry (frame) 
     self.price.grid(row = 2, column = 2) 

     ttk.Button (frame, text= 'Add record').grid (row = 3, column =2) 
     self.message = Label (text = '',fg = 'red') 
     self.message.grid (row = 3, column = 0) 

     self.tree = ttk.Treeview (height = 10, colum =2) 
     self.tree.grid(row = 4, column = 0, columnspan = 2) 
     self.tree.heading('#0', text = 'Name', anchor = W) 
     self.tree.heading(2, text = 'Price', anchor = W) 

     ttk.Button (text = 'Delete record').grid (row = 5, column = 0) 
     ttk.Button (text = 'Edit record').grid (row = 5, column = 1) 

     self.viewing_records() 



    def run_query (self, query, parameters =()): # database connection 
     with sqlite3.connect(self.db_name) as conn: 
      cursor = conn.cursor() 
      query_result = cursor.execute (query, parameters) 
      conn.commit() 
     return query_result 


    def viewing_records(self): 
     records = self.tree.get_children() 
     for element in records: 
      self.tree.delete (element) 
     query = 'SELECT * FROM product ORDER BY name DESC' 
     db_rows = self.run_query (query) 
     for row in db_rows: 
      self.tree.insert ('', 0, text = row[1], values = row[2]) 

if __name__ == '__main__': 
    wind = Tk() 
    application = Product (wind) 
    wind.mainloop() 

Answer 1

這將創建一個名爲db_path變量，它包含文件database.db

db_path = os.path.join(BASE_DIR, "database.db") 

的路徑，但這裏使用了文字字符串db_path作爲數據庫的名稱來連接，不的剛剛創建的變量的內容：

with sqlite3.connect('db_path') as conn: 

要使用變量，去掉引號：

with sqlite3.connect(db_path) as conn: 

Answer 2

... 
db_path = os.path.join(BASE_DIR, "database.db") 
with sqlite3.connect('db_path') as conn: 
... 

你正在創建一個名爲'db_path'一個新的數據庫文件。

而不是

with sqlite3.connect('db_path') as conn:

你應該有

with sqlite3.connect(db_path) as conn:

或者換句話說，使用變量db_path而不是文本字符串'db_path'。

爲了完整起見，你也可能要使用Product.db_name您已經定義的，而不是再硬編碼'database.db'：

... 
db_path = os.path.join(BASE_DIR, self.db_name) 
with sqlite3.connect(db_path) as conn: 
...