擅长:python、mysql、java
<p>除了<code>f.save(...)</code>定义之外,您所做的一切都是正确的</p>
<p>执行<code>f = uploaded_file.read()</code>时,<code>f</code>是<code>.read()</code>操作的结果,它是字节,而不是文件。
您必须打开另一个文件并将内容保存到其中。
不要忘记<code>.decode()</code>字节,使其成为字符串</p>
<p>下面是一个工作片段:</p>
<pre class="lang-py prettyprint-override"><code>
@app.route('/', methods=['POST'])
def upload_files():
uploaded_file = request.files['file']
filename = secure_filename(uploaded_file.filename)
if filename != '':
uploaded_file.stream.seek(0)
f = uploaded_file.read().decode()
# WE don't want any failures
f = f.replace("FAIL", "SUCCESS")
filename_to_save = os.path.join(app.config['UPLOAD_PATH'], filename)
with open(filename_to_save, "w") as file_to_save:
file_to_save.write(f)
return {"status": "OK"}
</code></pre>