> Hello. Newbie to django and first-time poster to this group.
welcome!
> class Employee(models.Model):
> name = models.CharField(maxlength=50)
> office = models.CharField(maxlength=20)
> phone= models.CharField(maxlength=15)
>
> class Photo(models.Model):
> filename = models.CharField(maxlength=50)
> master_photo = models.BooleanField()
> employee = models.ForeignKey(Employee)
>
> In one of my views I'd like to have a table consisting of rows like:
>
> <td><Photo.filename></td><td><Employee.name></
> td><td><Employee.office></td>
>
> but if I say:
>
> object_list = Employee.objects.order_by('name')
I think you want something like
object_list = Photo.objects.select_related().order_by(SORT_FIELD)
and then in your template you want
{% for photo in object_list %}
<tr>
<td>photo.filename</td>
<td>photo.employee.name</td>
<td>photo.employee.office</td>
</tr>
{% endfor %}
I never remember what the value for SORT_FIELD should be off the
top of my head...so you'd have to try some combination of your
app-name, object-name, and field-name, joined by either
underscores or periods. My first attempt would be
SORT_FIELD = 'app_employee.name'
though it might be something like one of these
SORT_FIELD = 'app_employee__name'
SORT_FIELD = 'employee.name'
SORT_FIELD = 'employee__name'
where "app" is the name of the app in your project.
Alternatively, you might be able to do something like
object_list = Employee.objects.select_related().order_by('name')
and then use
{% for emp in object_list %}
{% for photo in emp.photo_set %}
<tr>
<td>photo.filename</td>
<td>emp.name</td>
<td>emp.office</td>
</tr>
{% endfor %}
{% endfor %}
or even
{% for emp in object_list %}
<tr>
<td>emp.name</td>
<td>emp.office</td>
<td>
<ul>
{% for photo in emp.photo_set %}
<li>photo.filename</li>
{% endfor %}
</ul>
</td>
</tr>
{% endfor %}
HTH,
-tim
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