如何将输出文件保存在新文件夹中？现在这些存储在根目录中。指导我保存在特定位置，我想给出文件名

Question

请帮我升级以下内容：这实际上是打开网站并截图。

我需要将 o/p 文件保存在所需位置。如果可能的话，请帮我取名字。

示例：如果输入为：https://www.amazon.in/B078BNQ318/ 我希望将文件保存为 B078BNQ318。请帮忙

import webbrowser
import pyautogui, time
import xlrd

URL = []
chrome_path = 'C:/Program Files (x86)/Google/Chrome/Application/chrome.exe %s'

# change the loc as per your excel location
loc = ("C:\\Users\\dilip\\Documents\\URL.xlsx")
wb = xlrd.open_workbook(loc)
# open the first sheet
sheet = wb.sheet_by_index(0)

for i in range(sheet.nrows):
    URL.append(sheet.cell_value(i, 0))

for i in range(0, len(URL)):
    webbrowser.get(chrome_path).open(URL[i])
    time.sleep(5)
    screenshot = pyautogui.screenshot()
    filename = "file" + str(i) + ".png"
    screenshot.save(filename)

Answer 1

我确定您可以在路径上使用 split 并获取最后一个元素。 代码我看不懂，就说我懂的吧

#assume this is the url

url = 'https://www.amazon.in/B078BNQ318/'
i = url.split('/')

#assuming all url's are like the example you gave 

filename = "file" + str(i[-2]) + ".png"

#output of i = ['https:', '', 'www.amazon.in', 'B078BNQ318','']

Answer 2

你可以在python中使用f字符串。

for i in range(0, len(URL)):
    webbrowser.get(chrome_path).open(URL[i])
    time.sleep(5)
    screenshot = pyautogui.screenshot()
    filename = "file" + str(i) + ".png"
    screenshot.save(f"D:\\{filename}")   # Here I gave D drive as location.Change this acccording to your need