How to get the latest value of one filed(Django Model.objects.filter())

Question

I have models as below:

class ProjectRecord(models.Model):
    user = models.ForeignKey(User,on_delete=models.CASCADE, null=True, blank=True,related_name='user_projects')
    project = models.ForeignKey(Project,on_delete=models.CASCADE, null=True, blank=True, related_name='user_projects')
    project_type = models.CharField(max_length=30, choices=(("project_a","A"),("project_b","B")),null=True)
    version = models.FloatField(null=True, blank=True)

I want to filter the latest value of version with this: project = list(ProjectRecord.objects.filter(user=self.request.user, project_type='project_a')) , but I don't know how to achieve it.the data in database is similar as below:

id project_id version project_type
1   5          1.0      project_a
2   5          1.0      project_b
3   4          1.0      project_a
4   4          1.0      project_b
5   5          2.0      project_a          
6   5          2.0      project_b
7   5          3.0      project_a
8   5          3.0      project_b

For example, I want to get the latest value of project_id=5 to exact match other data and do not delete other project_id's value if their versions are not updated, the queryset should be display as below

id project_id version project_type
1   4          1.0      project_a
2   4          1.0      project_b
3   5          3.0      project_a
4   5          3.0      project_b

Thanks so much for any advice and assistance.

score 1 · Answer 1 · answered Jun 09 '18 at 15:14

1

Or you can try with this:

from django.db.models import Max

queryset = ProjectRecord.objects.filter(user=self.request.user) \
                                .values('project_id', 'project_type') \
                                .annotate(max_version=Max('version')

answered Jun 09 '18 at 15:14

Hi, I would like to use get_Foo_display in template, because project_type are choices, I need to display the detail of project_type, can I use other method to select Max version, because I can't use .values() and get_FOO_display in the meantime...... – Elsa Jul 02 '18 at 07:49

score 0 · Answer 2 · answered Jun 08 '18 at 08:17

0

You could order_by('version') and then take latest object and watch if the second one have the same ver. so you take and it in and the third ...

answered Jun 08 '18 at 08:17

Ivan Lavrenov

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7

1

Hi Ivan, thanks for your advice, I think order_by() is just a sort, values should be also display on pages. I want to display the latest version according to project ID, Thanks all the same. – Elsa Jun 09 '18 at 05:07

score 0 · Accepted Answer · answered Jun 08 '18 at 08:37

0

Try this,

from django.db.models import Max

queryset = ProjectRecord.objects.filter(user=self.request.user, project_type='project_a')
max_version = queryset.aggregate(max=Max('version')).get('max')
required_queryset = queryset.filter(version=max_version)

I'm not sure about the performance, but this will definitely work

answered Jun 08 '18 at 08:37

JPG

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Hi Jerin, thanks for your advice, I got a max value of project, but it lost something important values too, I also need other project_id's values, if their versions are not updated. Thanks again for your assistance, do you have any idea about this? – Elsa Jun 09 '18 at 05:05
hi peter, I would like to use `get_Foo_display` in template, because project_type are choices, I need to display the detail of `project_type`, can I use other method to select Max version, because I can't use .values() and get_FOO_display in the meantime...... – Elsa Jul 02 '18 at 07:49
I don't think it would be easy, anyway check this [SO answer](https://stackoverflow.com/a/21530804/8283848) – JPG Jul 02 '18 at 08:08
Still don't know how to use @property in my project, thank you all the same, I just would like to display the detail in template of `project_type`, it really doesn't have a simpler way to solving this? – Elsa Jul 02 '18 at 08:21

How to get the latest value of one filed(Django Model.objects.filter())

3 Answers3