Skip to Content
Menu
Dette spørgsmål er blevet anmeldt
2 Besvarelser
4632 Visninger

I want to the user to upload an excel file using an HTML form (input type="file") and then process it in a controller and make records based on its rows. how can I achieve it? Thanks in advance.

Avatar
Kassér
Forfatter Bedste svar
  1. I've figured it out myself.

xml:

<form class="o_form_binary_form" method="POST" enctype="multipart/form-data" action="/my/order/excel/uploaded">

    <input type="hidden" name="csrf_token" t-att-value="request.csrf_token()"/>

    <label class="control-label" for="file">File:</label>

    <input accept="xls, .xlsx" class="o_input_file" type="file" name="file"/>

    <button type="submit" class="btn btn-primary pull-left">Submit</button>

</form>


python controller:

from xlrd import open_workbook
@http.route(['/my/order/excel/uploaded'], type='http', auth="user", website=True)
def portal_create_order_excel(self, file):
    excel_data = file.read()
    book = open_workbook(file_contents=excel_data)
    sheet = book.sheets()[0]
     for i in range(sheet.nrows):
        // do what you do



Avatar
Kassér
Bedste svar

Hi you can check this: 


Avatar
Kassér
Related Posts Besvarelser Visninger Aktivitet
0
feb. 21
2554
3
nov. 20
4931
1
jul. 20
15953
1
jan. 21
3294
2
okt. 24
3743