使用python从Excel工作表中提取图像

Question

我找到了一些 Python2 代码来从 Excel 文件中提取图像。

我有一个非常基本的问题：我应该在哪里指定目标 excel 文件的路径？

或者它只适用于活动的打开的 Excel 文件？

import win32com.client       # Need pywin32 from pip
from PIL import ImageGrab    # Need PIL as well
import os

excel = win32com.client.Dispatch("Excel.Application")
workbook = excel.ActiveWorkbook

wb_folder = workbook.Path
wb_name = workbook.Name
wb_path = os.path.join(wb_folder, wb_name)

#print "Extracting images from %s" % wb_path
print("Extracting images from", wb_path)

image_no = 0

for sheet in workbook.Worksheets:
    for n, shape in enumerate(sheet.Shapes):
        if shape.Name.startswith("Picture"):
            # Some debug output for console
            image_no += 1
            print("---- Image No. %07i ----", image_no)

            # Sequence number the pictures, if there's more than one
            num = "" if n == 0 else "_%03i" % n

            filename = sheet.Name + num + ".jpg"
            file_path = os.path.join (wb_folder, filename)

            #print "Saving as %s" % file_path    # Debug output
            print('Saving as ', file_path)

            shape.Copy() # Copies from Excel to Windows clipboard

            # Use PIL (python imaging library) to save from Windows clipboard
            # to a file
            image = ImageGrab.grabclipboard()
            image.save(file_path,'jpeg')

Answer 1

您可以像这样从现有的 Excel 文件中抓取图像：

from PIL import ImageGrab
import win32com.client as win32

excel = win32.gencache.EnsureDispatch('Excel.Application')
workbook = excel.Workbooks.Open(r'C:\Users\file.xlsx')

for sheet in workbook.Worksheets:
    for i, shape in enumerate(sheet.Shapes):
        if shape.Name.startswith('Picture'):  # or try 'Image'
            shape.Copy()
            image = ImageGrab.grabclipboard()
            image.save('{}.jpg'.format(i+1), 'jpeg')

Answer 2

xlsx 文件实际上是一个 zip 文件。 您可以直接从 xl/media 子文件夹中获取图像。 您可以使用 ZipFile 类在 python 中执行此操作。 您不需要有 MS Excel，甚至不需要在 Windows 中运行！

Answer 3

文件路径和文件名在此处的变量中定义：

wb_folder = workbook.Path
wb_name = workbook.Name
wb_path = os.path.join(wb_folder, wb_name)

在这种特殊情况下，它在前面的行调用活动工作簿：

workbook = excel.ActiveWorkbook

但理论上您应该能够使用wb_folder和wb_name变量指定路径，只要您将文件加载到 excel 模块（ Python: Open Excel Workbook using Win32 COM Api ）。