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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()
701
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()
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