SimpleHTTPServerWithUpload

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an arbitrary file upload vulnerability

1

Based on the description of [this article](https://0day.work/finding-an-arbitrary-file-upload-vulnerability-in-a-filesharing-script/) , it is suggested that the parameter re.findall() at line 127 be changed.Change to: `fn = re.findall(r'Content-Disposition.*name="file"; filename="([^\/]*)"', line)`

lomo5

this is my web project

Anisha7320

Thank u. Please see the little change.

chrisju