用python设计一个图片浏览器，点击“上一张”按钮，则显示前一张图片，点击“下一张”按钮， 则显示后一张图片。

设计一个简单的图片浏览器功能，我们可以使用Python的Tkinter库来创建GUI界面，以及os库来处理文件路径。以下是基本的步骤和代码示例：

首先安装必要的库（如果还没有安装）：

pip install tk

然后，你可以编写如下的Python代码：

import os
from tkinter import *
from tkinter import filedialog

# 初始化变量
current_image_index = 0
image_path_list = []

def browse_images():
    global image_path_list
    directory = filedialog.askdirectory()
    if directory:
        images_in_dir = [f for f in os.listdir(directory) if f.endswith('.jpg') or f.endswith('.png')]
        image_path_list = [os.path.join(directory, img) for img in images_in_dir]
        display_image()

def display_image():
    global current_image_index
    try:
        with Image.open(image_path_list[current_image_index]) as im:
            canvas.delete("all")
            canvas.create_image(0, 0, anchor=NW, image=im)
            next_button.config(state="normal" if current_image_index < len(image_path_list) - 1 else "disabled")
            prev_button.config(state="normal" if current_image_index > 0 else "disabled")
            current_image_index += 1 if next_button["state"] == "normal" else -1
    except Exception as e:
        print(f"Error opening image: {e}")

root = Tk()
root.title("图片浏览器")

canvas = Canvas(root, width=400, height=400)
canvas.pack()

browse_button = Button(text="选择图片目录", command=browse_images)
browse_button.pack()

prev_button = Button(text="<< 上一张", state="disabled", command=lambda: display_image() if current_image_index > 0 else None)
prev_button.pack(side=LEFT)

next_button = Button(text="下一张 >", state="disabled", command=lambda: display_image() if current_image_index < len(image_path_list) - 1 else None)
next_button.pack(side=RIGHT)

root.mainloop()