Re: with django, how to write a grdview or table to show information and add a button or link in each row for approval?

i write with this method, but nothing is displayed, only blank page,

however there are data shown in console

where is wrong?

{% block content %}

    <table>

    {% for post in posts %}

      <tr>

        <td>

          post.project

        </td>

        <td>

          post.company

        </td>

      </tr>

    {% endfor %}

    </table>

{% endblock %}

views.py

def post_detail(request):

    posts = []

    with open('/home/martin/Downloads/site1/site1/reg/newusers.csv', 'rb') as csvfile:

     reader = csv.DictReader(csvfile)

     for row in reader:

      if len(row['name'].strip()) > 0 and len(row['project'].strip()) > 0 and len(row['initial password'].strip()) > 0 and len(row['email address'].strip()) > 0 and len(row['company'].strip()) > 0 and len(row['userType'].strip()) > 0 and len(row['businessType'].strip()) > 0 and len(row['contact'].strip()) > 0 and len(row['name'].strip()) > 0:

       eachbracket = {'name' : row['name'].strip(), 'project' : row['project'].strip(), 'businessType' : row['businessType'].strip(), 'userType' : row['userType'].strip(), 'company' : row['company'].strip()}

       posts = posts + [eachbracket];

    print posts

    return render(request, 'approval.html')

On Friday, May 27, 2016 at 3:25:35 PM UTC+8, jorr...@gmail.com wrote:

Sounds like Formsets is what you are looking for.

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