Re: Login

Check this: https://simpleisbetterthancomplex.com/tutorial/2016/06/27/how-to-use-djangos-built-in-login-system.html El lunes, 4 de diciembre de 2017, 14:49:32 (UTC-3), chaitanya.creator escribió: > > Hi > i am new users django .please help in making a login system using django > Thank you >

Update dict JsonField Postgresql

Hi everyone, I have a model model like this: class ItemCampaign(models.Model): campaign = models.ForeignKey( Campaign, related_name="itemscampaign", verbose_name="Item campaña" ) data = JSONField(default=dict) def __str__(self): return self.campaign.name The field dat

Re: Update dict JsonField Postgresql

e: >>> Model.objects.create(data={[..my dicts ]} That's wrong? Thanks! El lunes, 27 de marzo de 2017, 11:52:37 (UTC-2), Melvyn Sopacua escribió: > > On Monday 27 March 2017 05:56:34 Martin Peveri wrote: > > > > > But not working. > > > > >

Re: Update dict JsonField Postgresql

ocessed': False}, { 'number': '1162341721', 'id': 4, 'content': 'hello', 'processed':False}, { 'number': '1162341721', 'id': 5, 'content': 'hello', 'processed': False}]

Re: Update dict JsonField Postgresql

et all the records: objectitem.filter(data__contains=[{'processed': True}]) Thanks!!! El lunes, 27 de marzo de 2017, 14:29:13 (UTC-3), Melvyn Sopacua escribió: > > Hi, > > > > On Monday 27 March 2017 08:50:04 Martin Peveri wrote: >

Re: Update dict JsonField Postgresql

tes, 28 de marzo de 2017, 6:58:51 (UTC-3), Melvyn Sopacua escribió: > > On Monday 27 March 2017 15:03:17 Martin Peveri wrote: > > > Perfect!! This code works: > > > > > > data = objectitem.data > > > > > > data[0]['processed']

Filter list dict in JsonField Postgres

Hi, I have this model: class ItemCampaign(models.Model): campaign = models.ForeignKey( Campaign, related_name="itemscampaign", verbose_name="Item campaña" ) data = JSONField(default=dict) def __str__(self): return self.campaign.name With a record with this data: [{'