Django iteration (for ... in ...) how to put it in different divs? - django

Question

I have few instances of model.
my model:
class Record(models.Model):
name = models.ForeignKey(Car)
image = models.ImageField(upload_to='images/')
created = models.DateTimeField(
default=timezone.now)
view:
def allrecords(request):
records = Record.objects.all().order_by('created')
return render(request, 'mycar/allrecords.html', {'records' : records})
I want show it on my website. In my template i have:
{% for record in records %}
<img src={{ record.image.url}}/>
<div>
{{record.name}}
</div>
{% endfor %}
Now i get list of my records, but i would like put the newest record to first div, next to second etc. How can i do that?
I show simple screen how i would like have that (if someone will create new record, it will go to first div and other records will change place. Is any possibility to do something like that?
edit:
<div>
{% for record in records %}
{% if forloop.counter == 1 %}
<img src={{ record.image.url}}/>
<div>
{{record.name}}
</div>
{% endif %}
{% endfor %}
</div>
<div>
{% for record in records %}
{% if forloop.counter == 2 %}
<img src={{ record.image.url}}/>
<div>
{{record.name}}
</div>
{% endif %}
{% endfor %}
</div>
.
.
# till your 5th image

Answer 1

You can use forloop.counter to get the iteration number and check what is the iteration the loop and handle data accordingly.
In addition you can use CSS to make the layout work as you want.
Here is the information for Django template counter
Edit :
{% for record in records %}
<div>
{% if forloop.counter == 1 %}
# Here you can get your first images
<img src={{ record.image.url}}/>
<div>
{{record.name}}
</div>
{% endif %}
</div>
<div>
{% if forloop.counter == 2 %}
# Here you can get your first images
<img src={{ record.image.url}}/>
<div>
{{record.name}}
</div>
{% endif %}
</div>
.
.
# till your 5th image
{% endfor %}

Answer 2

There are two ways to do this. If you want to set this option for a single view then:
def all_records(request):
records = Record.objects.all().order_by('-created')
return render(request, 'mycar/allrecords.html', {'records' : records})
You're almost correct but order_by('created') leads to asceding order while order_by('-created') leads to descending order which is what you require.
Alternatively, if you want to have this setting to apply to all views then set class Meta in your models.py which will ensure that wherever you use Record.objects.all() it returns Records in descending order of created field:
class Record(models.Model):
name = models.ForeignKey(Car)
image = models.ImageField(upload_to='images/')
created = models.DateTimeField(
default=timezone.now)
class Meta:
ordering = ('-created')
It's Django design pattern to make all logical decisions in models and views and only just plugin formatted data in templates. You shouldn't add any complex logic in templates.

Answer 3

I'm assuming the question means that the model might have more than 5 records. If so, a more generic solution would be
<div class='row'>
<div class='firstimage'>
<img src={{ records[0].image.url}}/>
{{record.name}}
</div>
{% for record in records %}
{% if forloop.counter > 1 %}
<div class='subsequentimage'>
<img src={{ record.image.url}}/>
{{record.name}}
</div>
{% endif %}
{% cycle "" "</div><div class='row'>" "" %}
{% endfor %}
</div>
Note the use of the 'cycle' tag to begin a new row div every third cell div.
I don't know what your CSS classes are to distinguish between rows and cells so I used 'row', 'firstimage' (which might be defined to take up twice as much width) and 'subsequentimage' as example classes.

Answer 4

I recommend you to use the context variables:
def all_records(request):
records = Record.objects.all().order_by('-created')
newest = records[:5]
oldest = records[5:]
return render(request, 'mycar/allrecords.html', {'newst' : newest,
'oldest': oldest })
In your template :
{% for new in newst %}
<div>what you want with news</div>
{% endfor %}
{% for old in oldest %}
<div>what you want with olds</div>
{% endfor %}