Python

Question

我基本上是在尝试让 Python 在一个目录中创建一堆文件夹，每个文件夹的名称都基于 Excel 文件中的列表。该列表位于 D 列，标题为 "Folder Name".

我已经能够对单个单元格执行此操作，但很难弄清楚如何对多个单元格执行此操作。我到目前为止的代码如下。

非常感谢您的帮助 - 我对此很陌生！`

import os
import openpyxl


def folder_creation(EXCEL_FILE_DIRECTORY, FOLDER_CREATION_LOCATION, EXCEL_FILE_NAME):
   os.chdir (EXCEL_FILE_DIRECTORY)
   workbook = openpyxl.load_workbook (EXCEL_FILE_NAME)
   sheet = workbook.get_sheet_by_name ('Sheet1')
   folderName = sheet ['D2'].value
   baseDir = FOLDER_CREATION_LOCATION 
   os.makedirs(os.path.join(baseDir, folderName))
   print ("\nFolder created in: ", os.path.join(baseDir, folderName))

Answer 1

for 循环？

import os
import openpyxl


def folder_creation(EXCEL_FILE_DIRECTORY, FOLDER_CREATION_LOCATION, EXCEL_FILE_NAME):
   os.chdir (EXCEL_FILE_DIRECTORY)
   workbook = openpyxl.load_workbook (EXCEL_FILE_NAME)
   sheet = workbook.get_sheet_by_name ('Sheet1')
   baseDir = FOLDER_CREATION_LOCATION 
   col = sheet['D']

   for cell in col:
        folderName = cell.value
        os.makedirs(os.path.join(baseDir, folderName))
        print ("\nFolder created in: ", os.path.join(baseDir, folderName))

Answer 2

您必须遍历所有列值。这对我有用（openpyxl v2.5）：

def folder_creation(EXCEL_FILE_DIRECTORY, FOLDER_CREATION_LOCATION,  EXCEL_FILE_NAME):
    os.chdir(EXCEL_FILE_DIRECTORY)
    workbook = openpyxl.load_workbook(EXCEL_FILE_NAME)
    sheet = workbook.get_sheet_by_name('Sheet1')

    col_values = [cell.value for col in sheet.iter_cols(
        min_row=2, max_row=None, min_col=4, max_col=4) for cell in col]

    for value in col_values:       
        folderName = value
        baseDir = FOLDER_CREATION_LOCATION
        os.makedirs(os.path.join(baseDir, folderName))
        print("\nFolder created in: ", os.path.join(baseDir, folderName))

Python - 从 excel 列列表创建多个文件夹

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