如何使用 python 中的 MySql 數據創建交互式列表

Question

我在我的 python 代碼中使用 Tkinter 而不是顯示站點中可用房間列表的代碼，它顯示一個列表，該列表使用Entry導致打印列表被更改而不是保持固定。 有沒有辦法我可以編輯代碼，以便它可以打印出按鈕列表我可以 select 打開那個房間的問題。

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

Answer 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))

這是我用於 select 圖像並使用單選按鈕功能從表中刪除該行的代碼的一部分，您可以將刪除 function 替換為您選擇使用所選按鈕執行的任何操作。

Answer 2

您可以創建按鈕而不是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()