如何用Python从ZIP文件提取CSV？下载解压遇BadZipFile错误

问题场景 #

你作为Python新手，尝试用以下代码从URL下载.zip文件并提取CSV，但运行后报错：

原代码 #

# importing necessary modules
import requests, zipfile
from io import BytesIO
print('Downloading started')

#Defining the zip file URL
url = 'https://enfxfr.dol.gov/data_catalog/MSHA/msha_accident_20230422.csv.zip'

# Split URL to get the file name
filename = url.split('/')[-1]

# Downloading the file by sending the request to the URL
req = requests.get(url)
print('Downloading Completed')

# extracting the zip file contents
zipfile= zipfile.ZipFile(BytesIO(req.content))
zipfile.extractall('C:/Users/ssawant/OneDrive - Iron Senergy/Desktop/MSHA converted data/msha_accident')

错误信息 #

Traceback (most recent call last):
  File "c:\Users\ssawant\OneDrive - Iron Senergy\Desktop\MSHA converted data\test_2.py", line 17, in <module>
    zipfile= zipfile.ZipFile(BytesIO(req.content))
  File "C:\Users\ssawant\AppData\Local\Programs\Python\Python39\lib\zipfile.py", line 1266, in __init__
    self._RealGetContents()
  File "C:\Users\ssawant\AppData\Local\Programs\Python\Python39\lib\zipfile.py", line 1333, in _RealGetContents
    raise BadZipFile("File is not a zip file")
zipfile.BadZipFile: File is not a zip file

问题原因 #

1. 变量名覆盖模块名：用zipfile作为变量名，直接覆盖了导入的zipfile模块，会干扰模块方法的调用，导致ZipFile对象初始化异常。
2. 未验证请求有效性：直接使用req.content，如果请求返回错误页面（如404、500状态），得到的内容并非有效Zip文件，触发BadZipFile错误。

修复方案 #

1. 修改冲突变量名 #

将存储ZipFile对象的变量名改为zip_ref或其他不与模块名重复的名称。

2. 添加请求状态校验 #

使用req.raise_for_status()检查请求是否成功，失败时直接抛出异常，避免处理无效内容。

3. 可选：先保存本地文件 #

若仍有问题，可先将下载内容保存为本地文件，方便排查是否为有效Zip。

修复后的完整代码 #

import requests
import zipfile
from io import BytesIO

print('Downloading started')
url = 'https://enfxfr.dol.gov/data_catalog/MSHA/msha_accident_20230422.csv.zip'
filename = url.split('/')[-1]

# 发送请求并校验状态
req = requests.get(url)
req.raise_for_status()  # 请求失败时直接抛出异常
print('Downloading Completed')

# 避免变量名覆盖模块
zip_ref = zipfile.ZipFile(BytesIO(req.content))
extract_path = 'C:/Users/ssawant/OneDrive - Iron Senergy/Desktop/MSHA converted data/msha_accident'
zip_ref.extractall(extract_path)
zip_ref.close()  # 关闭Zip文件对象
print(f"文件已成功解压到 {extract_path}")

额外排查建议 #

如果还是报错，先保存下载内容到本地排查：

# 请求成功后添加这段代码保存文件
with open(filename, 'wb') as f:
    f.write(req.content)

手动打开文件，若为HTML页面则说明URL失效或网站需要验证；若为损坏的Zip，尝试分块下载：

req = requests.get(url, stream=True)
req.raise_for_status()
with open(filename, 'wb') as f:
    for chunk in req.iter_content(chunk_size=8192):
        f.write(chunk)
# 从本地文件解压
zip_ref = zipfile.ZipFile(filename)
zip_ref.extractall(extract_path)
zip_ref.close()

内容的提问来源于stack exchange，提问作者So50935