用于文件上传的 Flask/Apache 提交按钮

Question

我有一个在 apache 后面运行的烧瓶应用程序，在我的 index.html 页面上我有一个文件上传按钮和一个提交按钮，如下所示：

<form id="package_form" action="" method="POST">
  <div>
    <p>Upload Packages:</p>
    <p><input id="upload_button" type="file" class="btn btn-default btn-xs"></p>
    <p><input id="submit_button" type="submit" class="btn btn-success" value="Upload">
  </div>
</form>

我希望它会发送 post 请求，flask 会捕获它并上传文件，如下所示：

from flask import render_template, request, Response, url_for
from app import app
from werkzeug import secure_filename

## uploading specs ##
UPLOAD_FOLDER = '/tmp/'
ALLOWED_EXTENSIONS = set(['deb'])

def allowed_file(filename):
    return '.' in filename and \
    filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS

## index page stuff ##
@app.route('/index', methods = ['GET', 'POST'])
def index():
## kerberos username
    secuser = request.environ.get('REMOTE_USER')

    user = { 'nick': secuser }


## file uploading stuff
if request.method == 'POST': 
    file = request.files['file']
    if file and allowed_file(file.filename):
        filename = secure_filename(file.filename)
        file.save(os.path.join(UPLOAD_FOLDER, filename))
        return redirect(url_for('/index',     
                        filename=filename))

## main return
return render_template("index.html",
    user = user)

上传文件按钮运行良好，一切正常，只是按下提交按钮时出现 400 错误，所以它必须是烧瓶方面的东西，但我不太确定可能是什么。

任何帮助将不胜感激:)

Answer 1

if request.method == 'POST':
    file = request.files['file']
    if file and allowed_file(file.filename):
        filename = secure_filename(file.filename)
        file.save(os.path.join("/tmp/", filename))

这就是诀窍！

虽然你还需要将这个添加到 index.html (name="file" 到upload_button)

<form id="package_form" action="" method="POST">
  <div>
    <p>Upload Packages:</p>
    <p><input id="upload_button" type="file" class="btn btn-default btn-xs" name="file"></p>
    <p><input id="submit_button" type="submit" class="btn btn-success" value="Upload">
  </div>
</form>

Answer 2

当您在表单中包含文件输入时，您需要将enctype="multipart/form-data" 添加到表单标签本身，以告诉浏览器以正确的格式发送内容。