Read file data without saving it in Flask

Question

I am writing my first flask application  I am dealing with file uploads  and basically what I want is to read the data content of the uploaded file without saving it and then print it on the resulting page  Yes  I am assuming that the user uploads a text file always   Here is the simple upload function i am using    app route   upload    methods   GET    POST    def upload        if request method     POST           file   request files  file           if file              filename   secure filename file filename              file save os path join app config  UPLOAD FOLDER    filename               a    file uploaded       return render template  upload html   data   a    Right now  I am saving the file  but what I need is that  a  variable to contain the content data of the file    any ideas

User · Answer

in function  def handleUpload        if  photo  in request files          photo   request files  photo           if photo filename                          image   request files  photo                 image string   base64 b64encode image read                image string   image string decode  utf-8                use this to remove b      to get raw string             return render template  handleUpload html  filestring   image string      return render template  upload html     in html file   lt html gt   lt head gt       lt title gt Simple file upload using Python Flask lt  title gt   lt  head gt   lt body gt         if filestring           lt h1 gt Raw image  lt  h1 gt         lt h1 gt   filestring   lt  h1 gt         lt img src  data image png base64    filestring    alt  alternate    gt          else           lt h1 gt  lt  h1 gt         endif     lt  body gt

User · Answer

If you want to use standard Flask stuff - there s no way to avoid saving a temporary file if the uploaded file size is   500kb  If it s smaller than 500kb - it will use  BytesIO   which stores the file content in memory  and if it s more than 500kb - it stores the contents in TemporaryFile    as stated in the werkzeug documentation   In both cases your script will block until the entirety of uploaded file is received   The easiest way to work around this that I have found is   1  Create your own file-like IO class where you do all the processing of the incoming data  2  In your script  override Request class with your own   class MyRequest  Request      def  get file stream  self  total content length  content type  filename None  content length None        return MyAwesomeIO  filename   w      3  Replace Flask s request class with your own   app request class   MyRequest   4  Go have some beer

User · Answer

In case we want to dump the in memory file to disk  This code can be used    if isinstanceof obj SpooledTemporaryFile       obj rollover

User · Answer

I was trying to do the exact same thing  open a text file  a CSV for Pandas actually    Don t want to make a copy of it  just want to open it   The form-WTF has a nice file browser  but then it opens the file and makes a temporary file  which it presents as a memory stream   With a little work under the hood    form   UploadForm     if form validate on submit           filename   secure filename form fileContents data filename          filestream    form fileContents data        filestream seek 0        ef   pd read csv  filestream          sr   pd DataFrame ef          return render template  dataframe html  tables  sr to html justify  center  classes  table table-bordered table-hover    titles    filename   form form

User · Answer

We simply did   import io from pathlib import Path      def test my upload self  accept json              Test my uploads endpoint for POST             data                  filePath       tmp bin                manifest      io StringIO str Path   file    parent                                               path to file npmlist json    read                                npmlist json                      headers                  a    A                b    B                    res   self client post api route for   test                                   data data                                 content type  multipart form-data                                  headers headers                                           assert res status code    200

User · Answer

FileStorage contains stream field  This object must extend IO or file object  so it must contain read and other similar methods  FileStorage also extend stream field object attributes  so you can just use file read   instead file stream read    Also you can use save argument with dst parameter as StringIO or other IO or file object to copy FileStorage stream to another IO or file object   See documentation  http   flask pocoo org docs api  flask Request files and http   werkzeug pocoo org docs datastructures  werkzeug datastructures FileStorage

User · Answer

I share my solution  assuming everything is already configured to connect to google bucket in flask   from google cloud import storage   app route   upload    methods   POST    def upload        if request method     POST             FileStorage object wrapper         file   request files  file                               if file              os environ  GOOGLE APPLICATION CREDENTIALS     app config  GOOGLE APPLICATION CREDENTIALS               bucket name    bucket name               storage client   storage Client               bucket   storage client bucket bucket name                Upload file to Google Bucket             blob   bucket blob file filename               blob upload from string file read      My post  Direct to Google Bucket in flask

[python] Read file data without saving it in Flask

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