Question

我只知道如何编写用于GIS的python。这个代码还有更多使用arcpy和地理处理工具....这只是我坚持的开始部分，因为我试图准备数据，所以我可以使用zipped文件夹中的shapefile来完成剩下的工作。我的剧本

我试图提示用户输入要搜索的目录。要使用此脚本，它将搜索压缩的zip文件，然后将所有文件解压缩到同一目录。

import zipfile, os


# ask what directory to search in 
mypath = input("Enter .zip folder path: ")
extension = ".zip"

os.chdir(mypath) # change directory from working dir to dir with files

for item in os.listdir(mypath):
    if item.endswith(extension):
        filename = os.path.abspath(item)
        zip_ref = zipfile.ZipFile(filename)
        zip_ref.extractall(mypath)
        zip_ref.close()

尝试了所有建议，但仍然存在以下问题：

import zipfile, os

mypath = input("Enter folder: ")


if os.path.isdir(mypath):
    for dirpath, dirname, filenames in os.listdir(mypath):
        for file in filenames:
            if file.endswith(".zip"):
                print(os.path.abspath(file))
                with zipfile.ZipFile(os.path.abspath(file)) as z:
                    z.extractall(mypath)

else:
    print("Directory does not exist.")

Answer 1

如果位置无效，您的第一个if-block将否定else-block。我会完全删除'else'运算符。如果你保留它，if-check有效地杀死程序。 “if folderExist”足以替换else。

import arcpy, zipfile, os


# ask what directory to search in 
folder = input("Where is the directory? ")
# set workspace as variable so can change location
arcpy.env.workspace = folder

# check if invalid entry - if bad, ask to input different location
if len(folder) == 0:
    print("Invalid location.")
    new_folder = input("Try another directory?")
    new_folder = folder
    # does the above replace old location and re set as directory location?



# check to see if folder exists
folderExist = arcpy.Exists(folder)
if folderExist:
    # loop through files in directory
        for item in folder:
        # check for .zip extension
            if item.endswith(".zip"):
                 file_name = os.path.abspath(item) # get full path of files
                 print(file_name)
                 zip_ref = zipfile.ZipFile(file_name) # create zipfile object
                 zip_ref.extractall(folder) # extract all to directory
                 zip_ref.close() # close file

如果您没有使用原始代码，这可能会更整洁：

import zipfile, os
from tkinter import filedialog as fd

# ask what directory to search in 
folder = fd.askdirectory(title="Where is the directory?")

# loop through files in directory
for item in os.listdir(folder):
     # check for .zip extension
     if zipfile.is_zipfile(item):
         file_name = os.path.abspath(item) # get full path of files
         # could string combine to ensure path
         # file_name = folder + "/" + item
         print(file_name)

         zip_ref = zipfile.ZipFile(file_name) # create zipfile object
         zip_ref.extractall(folder) # extract all to directory
         zip_ref.close() # close file

Answer 2

我不确定使用arcpy。然而...

要迭代目录中的条目，请使用os.listdir：

for entry_name in os.listdir(directory_path): 
    # ...

在循环内，entry_name将是directory_path目录中项目的名称。

当检查它是否以“.zip”结尾时，请记住比较是区分大小写的。使用str.lower时，您可以使用str.endswith有效地忽略大小写：

if entry_name.lower().endswith('.zip'):
        # ...

要获取条目的完整路径（在本例中为.zip），请使用os.path.join：

entry_path = os.path.join(directory_path, entry_name)

将此完整路径传递给zipfile.ZipFile。

Answer 3

[增订]

我能够用以下代码解决相同问题

import os
import zipfile

mypath = raw_input('Enter Folder: ')

if os.path.isdir(mypath):
    for file in os.listdir(mypath):
        if file.endswith('.zip'):
            with zipfile.ZipFile(os.path.join(mypath, file)) as z:
                z.extractall(mypath)
else:
    print('Directory does not exist')

Python错误提取zip文件

3 个答案: