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的文件的名称。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

为什么不直接阅读有问题的文件并自行保存而不是提取？类似的东西：

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 2

您可以在每个文件被提取后立即运行重命名吗？ os.rename应该做到这一点。

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文件，只需开始计数，然后依次开启每个文件。

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

或者，如果您不关心结束名称，您可以随时使用类似uuid的内容。

import uuid
random_name = uuid.uuid4()

Answer 3

我怀疑在提取文件时可以重命名文件。提取文件后重命名文件怎么样？

依靠linux bash，你可以在一行中实现它：

os.system("find "+outpath+" -name '*.txt' -exec echo mv {} `echo {} | sed s/.txt/"+zipName+".txt/` \;")

因此，首先我们搜索指定文件夹中的所有txt文件，然后执行重命名命令，并使用sed计算新名称。

代码未经过测试，我现在正在使用Windows ^^＆＃39;

Answer 4

outpath = '/SCRATCH/emaxx'
suffix = os.path.splitext(item)[0]

for name in filelist :
    index = zip_ref.namelist().find(name)
    if index != -1: # check the file exists in the zipfile
        filename, ext = os.path.splitext(name)
        zip_ref.filelist[index].filename = f'{filename}_{suffix}.{ext}' # rename the extracting file to the suffix file name
        zip_ref.extract(zip_ref.filelist[index], outpath) # use the renamed file descriptor to extract the file

Answer 5

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/

从zipfile重命名解压缩的文件

5 个答案: