Django - cannot make url with variable in the template

Question

I want to have a button which links to page make-appointment

template.html

{% for key, value in closest_free_spots.items %}
...
<a href="{% url 'make-appointment' value.pk %}"><button type="button">{{ value.datetime }}</button></a>

urls.py

    url(r'^umowienie-spotkania/(?P<pk>\d+)/$', TestPy.views.MakingAppointmentView.as_view(),
            name='make-appointment'),

However when I try to launch it, I get error:

Reverse for 'make-appointment' with arguments '('',)' not found. 1 pattern(s) tried: ['umowienie-spotkania/(?P<pk>\\d+)/$']

It looks like the value.pk is not correct, but when I print it one line above, it gives correct value (i.e. 4).

What is the problem? What I'm missing?

closest_free_spots definition - value is object from CalendarFreeSlot model

    closest_free_spots = {
        lawyer: TestPy.models.CalendarFreeSlot.objects.filter(Q(lawyer_id=lawyer) & Q(is_available=True)).first() for
        lawyer in object_list}

models.py

class CalendarFreeSlot(models.Model):
    lawyer_id = models.ForeignKey('MyUser', on_delete=models.PROTECT) 
    datetime = models.DateTimeField()
    is_available = models.BooleanField(default=True)

Alasdair · Accepted Answer · 2019-09-05 12:49:49Z

1

Your closest_free_spots dict comprehension uses CalendarFreeSlot.objects.filter(...).first(), which can return None. This leads to the with arguments '('',)' part of the error message.

Checking {{ value.pk }} in the template was a good idea, but it looks like you missed that there isn't a value for every item in the dictionary.

You could handle this by checking {% if value %} in the template.

{% for key, value in closest_free_spots.items %}
{% if value %}
<a href="{% url 'make-appointment' value.pk %}"><button type="button">{{ value.datetime }}</button></a>
{% else %}
{{ key }} has not free spots
{% endif %}
{% endfor %}

answered Sep 5, 2019 at 12:49

Alasdair

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