Read File Data Without Saving It in Flask

Read file data without saving it in Flask

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.

How to read uploaded file data without saving it in Flask?

If you want only to read data from file then simply use .read()

uploaded_file = request.files.get('uploaded_file')

data = uploaded_file.read()

print(data)

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And if you want to use it with function which needs filename then usually it may work also with file-like object and you can use it directly

ie. Pillo.Image

from PIL import Image

uploaded_file = request.files.get('uploaded_file')

img = Image.open(uploaded_file)
img.save('new_name.jpg')

ie. pandas.DataFrame

import pandas as pd

uploaded_file = request.files.get('uploaded_file')

df = pd.read_csv(uploaded_file)

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You can also use io.BytesIO() to create file-like object in memory.

from PIL import Image
import io

uploaded_file = request.files.get('uploaded_file')

data = uploaded_file.read()

file_object = io.BytesIO(data)
#file_object.seek(0)

img = Image.open(file_object)
img.save('new_name.jpg')

import pandas as pd
import io

uploaded_file = request.files.get('uploaded_file')

data = uploaded_file.read()

file_object = io.BytesIO(data)
#file_object.seek(0)

df = pd.read_csv(file_object)

But this is more useful when you want to save/generate data without writing on disk and later send it back to browser.

uploaded_file = request.files.get('uploaded_file')

img = Image.open(uploaded_file)

# generate smaller version
img.thumbnail((200,200))

# write in file-like object as JPG
file_object = io.BytesIO()
img.save(file_object, 'JPEG')

# get data from file
data = file_object.getvalue()  # get data directly
# OR
#file_object.seek(0)           # move to the beginning of file after previous writing/reading
#data = file_object.read()     # read like from normal file

# send to browser as HTML
data = base64.b64encode(data).decode()
html = f'<img src="data:image/jpg;base64, {data}">'

Python flask upload file but do not save and use

I know that this is very outdated but for the sake of people landing here for similar inquiry, here it is if you want to save AND read your file after wards. Seems like Werkzeug's FileStorage class (which is the class to handle uploaded file in Flask) pointing to end of file after every action (saving or reading). So we have to move the pointer up to the beginning of the file before doing any subsequent action. I am using python's pandas in my answer below because I usually read csv into dataframe.

import pandas as pd

@app.route('/', methods=['GET', 'POST'])
    def upload_file():
        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(app.config['UPLOAD_FOLDER'], filename))

                ## snippet to read code below
                file.stream.seek(0) # seek to the beginning of file
                myfile = file.file # will point to tempfile itself
                dataframe = pd.read_csv(myfile)
                ## end snippet

                return "yatta"
            else:
                return "file not allowed"

        return render_template("index.html")

Flask upload file without saving it

This simplified code works in my app.

import tempfile
tempdirectory = tempfile.gettempdir()

class UploadView(Roled, BaseView):
if request.method == 'POST':
    if request.form['action'] == 'Upload':
        file = request.files['newfile']
        filename = secure_filename(file.filename)
        file.save(os.path.join(tempdirectory, filename))

and in the template:

<form action="" method="POST" enctype="multipart/form-data">
     <div>
       <input type="file" name="file"/>
     </div>
    <input class="btn" type="submit" name="action" value="Upload">
</form>

How to parse csv into dictionary in Python without saving file?

Your error says your file is encoded, decode it.

fstring = f.read().decode("utf8")

Opening a file that has been uploaded in Flask

Answering my own question in case someone else needs this.

FileStorage objects have a .stream attribute which will be an io.BytesIO

f  = request.files['data_file']
df = pandas.read_csv(f.stream)

Return File to client side without save in sever side which download from GCP cloud storage

Base on JohnHanley's comments able to achieve the goal with the below code. from that way can serve any files without concern about the content type

@app.route('/api/download-file', methods=['GET'])
@token_required
def download_blob():
    """Downloads a blob."""

    file_name = "dir/" + request.args.get('file_name')
    storage_client = storage.Client()

    bucket = storage_client.bucket(app.config.get('CLOUD_STORAGE_BUCKET'))
    blob = bucket.get_blob(file_name)
    content_type = None
    try:
        content_type = blob.content_type
    except:
        pass
    file = blob.download_as_string()
    print(type(file), "downloaded type")
    return Response(file,  mimetype=content_type)

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