从zipfile重命名解压缩的文件

Question

我在Linux服务器上有很多压缩文件,每个文件包含多个文本文件.

我想要的是提取一些文本文件,这些文件在压缩文件中具有相同的名称并将其保存为文件夹; 我正在为每个压缩文件创建一个文件夹,并将文本文件解压缩到它.我需要将父压缩文件夹名称添加到文件名的末尾,并将所有文本文件保存在一个目录中.例如,如果压缩文件夹是March132017.zip并且我提取了holding.txt,那么我的文件名将是holding_march13207.txt.

我的问题是我无法更改提取文件的名称.如果你能提出建议,我将不胜感激.

import os 
import sys 
import zipfile
os.chdir("/feeds/lipper/emaxx") 

pwkwd = "/feeds/lipper/emaxx" 

for item in os.listdir(pwkwd): # loop through items in dir
    if item.endswith(".zip"): # check for ".zip" extension
        file_name = os.path.abspath(item) # get full path of files
        fh = open(file_name, "rb")
        zip_ref = zipfile.ZipFile(fh)

        filelist = 'ISSUERS.TXT' , 'SECMAST.TXT' , 'FUND.TXT' , 'HOLDING.TXT'
        for name in filelist :
            try:
                outpath = "/SCRATCH/emaxx" + "/" + os.path.splitext(item)[0]
                zip_ref.extract(name, outpath)

            except KeyError:
                {}

        fh.close()

Answer 1

import zipfile

zipdata = zipfile.ZipFile('somefile.zip')
zipinfos = zipdata.infolist()

# iterate through each file
for zipinfo in zipinfos:
    # This will do the renaming
    zipinfo.filename = do_something_to(zipinfo.filename)
    zipdata.extract(zipinfo)

参考：https : //bitdrop.st0w.com/2010/07/23/python-extracting-a-file-from-a-zip-file-with-a-different-name/

Answer 2

为什么不阅读相关文件并自己保存而不是解压缩呢？就像是：

import os
import zipfile

source_dir = "/feeds/lipper/emaxx"  # folder with zip files
target_dir = "/SCRATCH/emaxx"  # folder to save the extracted files

# Are you sure your files names are capitalized in your zip files?
filelist = ['ISSUERS.TXT', 'SECMAST.TXT', 'FUND.TXT', 'HOLDING.TXT']

for item in os.listdir(source_dir):  # loop through items in dir
    if item.endswith(".zip"):  # check for ".zip" extension
        file_path = os.path.join(source_dir, item)  # get zip file path
        with zipfile.ZipFile(file_path) as zf:  # open the zip file
            for target_file in filelist:  # loop through the list of files to extract
                if target_file in zf.namelist():  # check if the file exists in the archive
                    # generate the desired output name:
                    target_name = os.path.splitext(target_file)[0] + "_" + os.path.splitext(file_path)[0] + ".txt"
                    target_path = os.path.join(target_dir, target_name)  # output path
                    with open(target_path, "w") as f:  # open the output path for writing
                        f.write(zf.read(target_file))  # save the contents of the file in it
                # next file from the list...
    # next zip file...

Answer 3

您可以在提取每个文件后简单地运行重命名，对吗？os.rename应该可以解决问题。

zip_ref.extract(name, outpath)
parent_zip = os.path.basename(os.path.dirname(outpath)) + ".zip"
new_file_name = os.path.splitext(os.path.basename(name))[0] # just the filename

new_name_path = os.path.dirname(outpath) + os.sep + new_file_name + "_" + parent_zip
os.rename(outpath, new_namepath)

对于文件名，如果您希望它是增量的，只需开始计数，然后对每个文件进行计数。

count = 0
for file in files:
    count += 1
    # ... Do our file actions
    new_file_name = original_file_name + "_" + str(count)
    # ...

或者，如果您不关心最终名称，则可以始终使用 uuid 之类的东西。

import uuid
random_name = uuid.uuid4()