我在Python 3.7中编写了这段代码
from tkinter import *
root = Tk()
import random
def cmd():
print("Files:", img1.cget("file"), img2.cget("file"))
img = random.choice([img1, img2])
can.itemconfig(log, image=img)
print("Choice:", can.itemcget(log, "image"))
img1 = PhotoImage(file="a.gif")
img2 = PhotoImage(file="b.gif")
can = Canvas(root, width=300, height=300, bg="white")
log = can.create_image(151, 151, image=None)
can.pack()
btt = Button(root, text="click", command=cmd)
btt.pack()
root.mainloop()
我想打印在画布内配置的图像的名称
print("Files:", img1.cget("file"), img2.cget("file"))
Result > Files: a.gif b.gif < Return Filename OK
print("Choice:", can.itemcget(log, "image"))
Result > Choice: pyimage1 Or Choice: pyimage2 < Return String No OK
我该怎么办?谢谢!
答案 0 :(得分:0)
但是无法通过相反的过程获取文件名吗?
示例:
print(PhotoImage(name="pyimage1"))
返回> a.gif