Upload, read, write excel file in Python flask

Question

Im using this code which asks user to upload a file, which I want to be read into a dataframe. Then this dataframe should be displayed as output on the page.

What should I write in the return, so as to accomplish this ?

from flask import Flask, request, jsonify
import flask_excel as excel
import pandas as pd

app=Flask(__name__)

@app.route("/upload", methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        return jsonify({"result": request.get_array(field_name='file')})
    return '''
    <!doctype html>
    <title>Upload an excel file</title>
    <h1>Excel file upload (csv, tsv, csvz, tsvz only)</h1>
    <form action="" method=post enctype=multipart/form-data>
    <p><input type=file name=file><input type=submit value=Upload>
    </form>
    '''

@app.route("/export", methods=['GET'])
def export_records():
    return 

if __name__ == "__main__":
    app.run()

Atul Vinayak · Accepted Answer · 2017-10-25 10:19:08Z

12

I guess a barebones version of what you wanted would be this. But this would obviously require more work.

from flask import Flask, request, jsonify
import pandas as pd

app=Flask(__name__)

@app.route("/upload", methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        print(request.files['file'])
        f = request.files['file']
        data_xls = pd.read_excel(f)
        return data_xls.to_html()
    return '''
    <!doctype html>
    <title>Upload an excel file</title>
    <h1>Excel file upload (csv, tsv, csvz, tsvz only)</h1>
    <form action="" method=post enctype=multipart/form-data>
    <p><input type=file name=file><input type=submit value=Upload>
    </form>
    '''

@app.route("/export", methods=['GET'])
def export_records():
    return 

if __name__ == "__main__":
    app.run()

answered Oct 25, 2017 at 10:19

Atul Vinayak

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