Question

我正在尝试让我的程序从文件（例如.txt）中读取名称列表，然后在选定的文件夹中搜索名称，然后将这些文件复制并粘贴到另一个选定的文件夹中。我的程序运行没有错误，但没有执行任何操作：

代码-已更新：

import os, shutil
from tkinter import filedialog
from tkinter import *


root = Tk()
root.withdraw()

filePath = filedialog.askopenfilename()
folderPath = filedialog.askdirectory()
destination = filedialog.askdirectory()

filesToFind = []
with open(filePath, "r") as fh:
    for row in fh:
        filesToFind.append(row.strip())

#Added the print statements below to check that things were feeding correctly
print(filesToFind)
print(folderPath)
print(destination)

#The issue seems to be with the copy loop below:    
for target in folderPath:
    if target in filesToFind:
        name = os.path.join(folderPath,target)
        print(name)
        if os.path.isfile(name):
            shutil.copy(name, destination)
        else:
            print ("file does not exist", name)
        print(name)

更新-运行无误，但不会移动任何文件。

Answer 1

程序的最后部分可能会通过以下方式更好地工作：

for file in files:
    if file in filesToFind:
        name = os.path.join( folderPath, file )
        if os.path.isfile( name ) :
            shutil.copy( name, destination)
        else :
            print 'file does not exist', name

否则，对于从当前文件夹中复制文件的位置，也许还有很多未知的地方，以及为什么不使用它需要更早输入folderPath。

btw，file是python中的保留字，我建议为您的变量使用另一个名称，该名称与python保留字不符。

Answer 2

您的上一节有问题。

for file in os.listdir(folderPath): 
    for file in files:
        if file in filesToFind:
            shutil.copy(file, destination)

第一个for循环遍历目录中的每个文件名，这是完全可以理解的。

第二个for是一个错误，因为files不存在。您打算怎么办？

Answer 3

有效的代码-

import os import shutil from tkinter import * from tkinter import filedialog

root = Tk() root.withdraw()

filePath = filedialog.askopenfilename() folderPath = filedialog.askdirectory() destination = filedialog.askdirectory()

# First, create a list and populate it with the files

# you want to find (1 file per row in myfiles.txt)

filesToFind = [] with open(filePath, "r") as fh: for row in fh: filesToFind.append(row.strip())

# Had an issue here but needed to define and then reference the filename variable itself for filename in os.listdir(folderPath): if filename in filesToFind: filename = os.path.join(folderPath, filename) shutil.copy(filename, destination) else: print("file does not exist: filename")

注意-需要在正在读取的文件中包含文件扩展名。感谢@lenik和@John Gordon的帮助！是时候对其进行完善以使其更加人性化了

Python-将列表中的特定文件复制到新文件夹

3 个答案: