瓶/ Apache的提交按钮上传文件 [英] Flask/Apache submit button for file upload

问题描述

我在后面跟着跑阿帕奇烧瓶中的应用，我的index.html页面上我有一个文件的上传按钮和一个提交按钮，在这里看到：

 ＆LT;表ID =package_form行动=方法=POST＆GT;
  ＆LT; D​​IV＆GT;
    ＆LT; P＆GT;上传软件包：其中; / P＆GT;
    ＆LT; P＆GT;＆LT;输入ID =upload_button类型=文件级=BTN BTN-默认BTN-XS＆GT;＆LT; / P＆GT;
    ＆LT; P＆GT;＆LT;输入ID =submit_button类型=提交级=BTN BTN-成功值=上传＆GT;
  ＆LT; / DIV＆GT;
＆LT; /表及GT;
 

我希望将帖子发送请求和烧瓶会赶上它，做文件上传如本文件：

 从瓶进口render_template，请求，响应url_for
从应用程序导入
从WERKZEUG进口secure_filename##载规格##
UPLOAD_FOLDER ='/ tmp目录/'
ALLOWED_EXTENSIONS =集（['DEB']）高清allowed_file（文件名）：
    回报'。在文件名和\\
    filename.rsplit（'。'，1）[1]在ALLOWED_EXTENSIONS##索引页的东西##
@ app.route（'/索引，方法= ['GET'，'POST']）
高清指数（）：
##名的Kerberos
    secuser = request.environ.get（'REMOTE_USER'）    用户= {'缺口'：secuser}
##文件上传的东西
如果request.method =='POST'：
    文件= request.files ['文件']
    如果文件和allowed_file（file.filename）：
        文件名= secure_filename（file.filename）
        file.save（os.path.join（UPLOAD_FOLDER，文件名））
        返回重定向（url_for（'/索引，
                        文件名=文件名））##主回
返回render_template（index.html的
    用户=用户）
 

上传文件按钮效果很好，正确做的一切，它只是，当提交按钮被pressed我得到一个400错误，所以它必须是对事物的烧瓶一面的东西，但我不完全相信它会是什么。

任何帮助将是非常美联社preciated：）

解决方案

 如果request.method =='POST'：
    文件= request.files ['文件']
    如果文件和allowed_file（file.filename）：
        文件名= secure_filename（file.filename）
        file.save（os.path.join（/ tmp目录/，文件名））
 

这做的伎俩！

虽然你还需要将它添加到index.html的（名字=文件到upload_button）

 ＆LT;表ID =package_form行动=方法=POST＆GT;
  ＆LT; D​​IV＆GT;
    ＆LT; P＆GT;上传软件包：其中; / P＆GT;
    ＆LT; P＆GT;＆LT;输入ID =upload_button类型=文件级=BTN BTN-默认BTN-XSNAME =文件＆GT;＆LT; / P＆GT;
    ＆LT; P＆GT;＆LT;输入ID =submit_button类型=提交级=BTN BTN-成功值=上传＆GT;
  ＆LT; / DIV＆GT;
＆LT; /表及GT;
 

I have a flask application running behind apache, and on my index.html page I have a file upload button and a submit button, as seen here:

<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>

which I was hoping would send the post request and flask would catch it and do the file uploading as shown in this file:

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)

The upload file button works well, and does everything correctly, it's just that when the Submit button is pressed I get a 400 error, so it has to be something on the flask side of things, but I'm not exactly sure what it could be.

Any help would be much appreciated :)

解决方案

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))

This does the trick!

Although you also need to add this to index.html (name="file" to the 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>

这篇关于瓶/ Apache的提交按钮上传文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！