如何在 python 中使用 MySql 数据创建交互式列表

Posted 2023-02-24

【中文标题】如何在 python 中使用 MySql 数据创建交互式列表

【英文标题】：How can I create an interactive list using MySql data in python

【发布时间】：2021-04-19 06:05:26

【问题描述】：

我在我的 python 代码中使用 Tkinter，而不是显示站点中可用房间列表的代码，而是显示一个使用 Entry 的列表，导致打印的列表被更改而不是保持固定。有没有办法可以编辑代码，以便打印出我可以选择为该房间打开问题的按钮列表。

sitename3_info = sitename.get()
# the label should print all of the rooms in the site that was inputted in the audit function
cursor = cnn.cursor()
# retrieves the siteID from the inputted site name
siteID_fetch2 = "SELECT siteID FROM Sites WHERE siteName = %s"
cursor.execute(siteID_fetch2, [sitename3_info])
siteID_fetch2 = cursor.fetchall()
print(siteID_fetch2[0][0])
# searches for all the rooms for the site that was inputted
room = "SELECT roomname FROM rooms WHERE siteID_fk2 = %s"
cursor.execute(room,[siteID_fetch2[0][0]])
printrooms = cursor.fetchall()
print(printrooms[0][0])
# prints out a list of rooms in the site
i=0
for rooms in printrooms:
    for j in range(len(rooms)):
        e = Entry(screen13, width=10, fg='blue')
        e.grid(row=i, column=j)
        e.insert(END, rooms[j])
    i=i+1

【参考方案1】：

    def change():
        global pos
        if j.get()==0:
            b2['state']='NORMAL'
            pos=0
        elif j.get()==1:
            b2['state']='NORMAL'
            pos=1
        elif j.get()==2:
            b2['state']='NORMAL'
            pos=2
        elif j.get()==3:
            b2['state']='NORMAL'
            pos=3
        elif j.get()==4:
            b2['state']='NORMAL'
            pos=4
    def dele():
        sql="""DELETE FROM myteam where Name = %s"""
        if pos==0:
          
            mycur.execute(sql,(name[0],))
        if pos==1:
            mycur.execute(sql,(name[1],))
        if pos==2:
            mycur.execute(sql,(name[2],))
        if pos==3:
            mycur.execute(sql,(name[3],))

        if pos==4:
            mycur.execute(sql,(name[4],))
        db.commit()
        messagebox.showinfo("ALERT","This player has been removed,select a new player")
        new.destroy()
    for i in range(0,len(name)):
        name[i]=name[i].strip("\'")
    for i in index:
        i = i.strip('\"')
        imgs.append(ImageTk.PhotoImage(Image.open(i)))
    j=IntVar()
    for i in range(0,len(index)):
        l2.append(Radiobutton(new,variable=j,value=i,image=imgs[i],state=ACTIVE,command = change))

这是我用来选择图像并使用单选按钮功能从表中删除该行的代码的一部分，您可以将删除功能替换为您选择对所选按钮执行的任何操作。

【参考方案2】：

您可以创建按钮而不是 Entry 小部件：

def action(room):
    print(room)
    # do whatever you want on the room

# assume OP code is inside a function
def search():
    sitename3_info = sitename.get().strip()
    if sitename3_info:
        with cnn.cursor() as cursor:
            # combine the two SQL statements into one
            sql = ("SELECT roomname FROM rooms, Sites "
                   "WHERE rooms.siteID_fk2 = Sites.siteID AND siteName = %s")
            cursor.execute(sql, [sitename3_info])
            rooms = cursor.fetchall()
        # remove previous result (assume screen13 contains only result)
        for w in screen13.winfo_children():
            w.destroy()
        if rooms:
            for i, row in enumerate(rooms):
                roomname = row[0]
                btn = Button(screen13, text=roomname, command=lambda room=roomname: action(room))
                btn.grid(row=i, column=0)
        else:
            Label(screen13, text="No room found").grid()

---

更新：不使用上下文管理器的示例：

def search():
    sitename3_info = sitename.get().strip()
    if sitename3_info:
        cursor = cnn.cursor()
        # combine the two SQL statements into one
        sql = ("SELECT roomname FROM rooms, Sites "
               "WHERE rooms.siteID_fk2 = Sites.siteID AND siteName = %s")
        cursor.execute(sql, [sitename3_info])
        rooms = cursor.fetchall()
        # remove previous result (assume screen13 contains only result)
        for w in screen13.winfo_children():
            w.destroy()
        if rooms:
            for i, row in enumerate(rooms):
                roomname = row[0]
                btn = Button(screen13, text=roomname, command=lambda room=roomname: action(room))
                btn.grid(row=i, column=0)
        else:
            Label(screen13, text="No room found").grid()

【讨论】：

我试图尝试您的方法，但是我收到错误：AttributeError: enter 我不确定我的代码需要更改什么或是否需要添加任何内容它。

以 cnn.connect() 作为光标：AttributeError: 'enter'

@Julia 那么您使用的 mysql 模块不支持上下文管理器功能。在不使用上下文管理器功能的情况下通过示例更新答案。