python - 使用python写入AWS Lambda中的/tmp目录

Question

目标

我正在尝试将zip文件写入python aws lambda中的/ tmp文件夹中，因此我可以在压缩之前提取操作，并将其放入s3存储桶中。

问题

Os Errno30只读文件系统

在我的计算机上对该代码进行了本地测试，以确保在将文件上传到AWS之前，该文件将写入我的工作目录。这是我要使用的代码。

file = downloadFile() #This is api call that returns binary zip object
newFile = open('/tmp/myZip.zip','wb')
newFile.write(file)
extractAll('/tmp/myZip.zip')

这是试图提取压缩文件的代码

def extractAll(self,source):
        with zipfile.ZipFile(source, 'r') as archive:
            archive.extractall()

这是痕迹

[Errno 30] Read-only file system: '/var/task/test-deploy': OSError
Traceback (most recent call last):
File "/var/task/web.py", line 31, in web
performAction(bb, eventBody)
File "/var/task/src/api/web.py", line 42, in performAction
zipHelper.extractAll('/tmp/myZip.zip')
File "/var/task/src/shared/utils/zipfilehelper.py", line 24, in extractAll
archive.extractall()
File "/var/lang/lib/python3.6/zipfile.py", line 1491, in extractall
self.extract(zipinfo, path, pwd)
File "/var/lang/lib/python3.6/zipfile.py", line 1479, in extract
return self._extract_member(member, path, pwd)
File "/var/lang/lib/python3.6/zipfile.py", line 1538, in _extract_member
os.mkdir(targetpath)
OSError: [Errno 30] Read-only file system: '/var/task/test-deploy'

Answer 1

extractAll()将提取当前目录中的文件，在您的情况下为/var/task/test-deploy。

您需要使用os.chdir()更改当前目录：

import os, zipfile

os.chdir('/tmp')
with zipfile.ZipFile(source, 'r') as archive:
    archive.extractall()

更好的是，您可以 create a temporary directory并在那里提取文件，以避免污染 /tmp：

import os, tempfile, zipfile

with tempfile.TemporaryDirectory() as tmpdir:
    os.chdir(tmpdir)
    with zipfile.ZipFile(source, 'r') as archive:
        archive.extractall()

如果要在提取文件后还原当前工作目录，请考虑使用此装饰器：

import os, tempfile, zipfile, contextlib

@contextlib.context_manager
def temporary_work_dir():
    old_work_dir = os.getcwd()
    with tempfile.TemporaryDirectory() as tmp_dir:
        os.chdir(tmp_dir)
        try:
            yield
        finally:
            os.chdir(old_work_dir)

with temporary_work_dir():
    with zipfile.ZipFile(source, 'r') as archive:
        archive.extractall()