在 Flask 中，如何临时存储用户提交的文件以对其进行操作并将其返回? [英] In Flask, how do I store a user submitted file temporarily to manipulate it and return it back?
问题描述
所以我一直在开发 Flask 应用程序，其中用户将文件提交到表单，服务器接收并操作它，然后将编辑后的文件返回给用户.我是 Flask 和网络开发的新手，所以请耐心等待.我已经使用表单和文件字段设置了网站，但我不确定如何从文件中获取数据.我必须将它存储在服务器上吗?如果我这样做了，我完成后是否必须删除?
So I've been working on a Flask application where the user submits a file to a form, the server takes it and manipulates it, then returns the edited file back to the user. I am new to Flask and web dev in general, so bear with me. I have already set up the website with the form and a filefield, but Im not exactly sure how I can get the data from the file. Do i have to store it on the server? If I do, do I have to delete after im done?
这是我目前的路线
@app.route('/mysite', methods=['post', 'get'])
def mysite():
form = mysiteform()
if request.method == 'POST' and form.validate():
file = form.file.data.read()
filename = secure_filename(file.filename)
return render_template('mysite.html', title = 'hello world!', form=form)
推荐答案
在 Flask 官方文档站点，有一个 文件上传示例如下:
In Flask official documentation site, there is a file uploader example as below:
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# check if the post request has the file part
if 'file' not in request.files:
flash('No file part')
return redirect(request.url)
file = request.files['file']
# If the user does not select a file, the browser submits an
# empty file without a filename.
if file.filename == '':
flash('No selected file')
return redirect(request.url)
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return redirect(url_for('download_file', name=filename))
return '''
<!doctype html>
<title>Upload new File</title>
<h1>Upload new File</h1>
<form method=post enctype=multipart/form-data>
<input type=file name=file>
<input type=submit value=Upload>
</form>
'''
您可以通过以下方式获取文件内容:file = request.files['file'] 当然您还可以对文件进行进一步操作.
Here is how you can get the file content: file = request.files['file'] and of course you can do further manipulation on the file.
以下是保存文件的方法file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename)).如果您想保存文件并稍后返回文件，或者只是在内存中修改它并立即返回，这完全取决于您.
Here is how you can save the file file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename)). It's really up to you if you want to save the file and return the file later or just modify it in memory and return it immediately.
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