属性错误：类型对象"图像"没有属性"打开"

Question

Exception in Tkinter callback
Traceback (most recent call last):
File "C:Python34libtkinter__init__.py", line 1482, in __call__
return self.func(*args)
File "C:UsersTAODesktopNewb.py", line 14, in imgs
img = Image.open("C:\Users\TAO\Desktop\New\02.png")
AttributeError: type object 'Image' has no attribute 'open'

当我运行代码时会出现此错误消息我正在尝试为注册系统和身份证打印机编码第一个表单使用sqlite3数据库和tkinter GUI将数据导入数据库。这个表单用于从数据库中获取数据并在照片数据上写入。制作身份证并打印它。但当我尝试使用image.open((打开图像文件时，它不起作用。

from PIL import *
import sqlite3
from tkinter import *
connection = sqlite3.connect("school.db")
tao = Tk()
tao.title("Mayurapada Central Collage")
tao.configure(bg = '#6699ff')
canvas = Canvas(tao,width = 600,height = 400,bg = '#6699ff')
def imgs():
img = Image.open("C:\Users\TAO\Desktop\New\02.png")
img.show()
str01 = "Image"
font = ImageFont.truetype("arial.ttf",75)
w,h = font.getsize(str01)
print(str01)

button01 = Button(tao,text = "Preview",bd = 7,padx = 5,pady = 5,command = 
imgs).place(x = 50,y = 300)
canvas.pack()
tao.mainloop()

Answer 1

问题是import *混淆了PIL.Image和tkinter.Image。

import sqlite3
from tkinter import Tk, Button, Canvas
from PIL import Image, ImageFont
connection = sqlite3.connect("school.db")
tao = Tk()
tao.title("Mayurapada Central Collage")
tao.configure(bg = '#6699ff')
canvas = Canvas(tao,width = 600,height = 400,bg = '#6699ff')
def imgs():
img = Image.open("C:\Users\TAO\Desktop\New\02.png")
img.show()
str01 = "Image"
font = ImageFont.truetype("arial.ttf",75)
w,h = font.getsize(str01)
print(str01)

button01 = Button(tao,text = "Preview",bd = 7,padx = 5,pady = 5,command = 
imgs).place(x = 50,y = 300)
canvas.pack()
tao.mainloop()