This is the code:
{% for o in page_obj.object_list %}
<tr style="color:#4A3C8C;background-color:{% cycle '#E7E7FF' '#F7F7F7' %};">
<td align="center"> {{o.terminal_id}}</td>
<td align="center"> {{o.time_stamp}}</td>
<td align="center" class="lat"> {{o.latitude|floatformat:"6"}}</td>
<td align="center" class="lng"> {{o.longitude|floatformat:"6"}}</td>
<td align="center"> {{o.speed}}</td>
<td align="center">
<script>
$("sc开发者_如何学Goript").last().parent().parent()
</script>
</td>
<td align="center"> {{o.speed}}</td>
<td align="center"> {{o.adress_reality}}</td>
</tr>
{%endfor%}
I want to get the next object after 'o'
object.
How do I get the next from the 'o'
object? Alternatively, how to get the next object in the python file that i can show this in the html as shown below?
{{ o.next_obj }}
thanks
If you want to access multiple objects within a forloop, (although this is not a good design idea, but that is a separate discussion entirely.) you don't loop on the objects, but a counter and access the respective objects' counters moved around.
#In your view
obj_count = range(page_obj.object_list.count())
{% for i in obj_count %}
o.i # Will access current object
o.i+1 # Will access the next object
{% endfor %}
did you try to use get_next_by_FOO ? https://docs.djangoproject.com/en/1.2/ref/models/instances/#django.db.models.Model.get_next_by_FOO
If the field is not generated on the fly, you could add to your model (naively):
class MyObj(models.Model):
lat = ...
lon = ...
def get_next_by_id(self):
return self.objects.get(id=self.id+1)
And use {{ o.get_next_by_id.lat }}
Otherwise it is a good use for the templatetags, see the tag section but require more code. If you provide me a more detailed example of what you are doing i could try to give you a generic templatetag which might work in your case.
Good luck.
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