Reputation: 139
I want to set a 'app_id'of application model as a foreign key for interviewstable's app_id
class application(models.Model):
app_id = models.IntegerField(max_length=200)
job_title = models.CharField(max_length=200)
odesk_id = models.CharField(max_length=200)
client_spent = models.CharField(max_length=200)
job_type = models.CharField(max_length=200)
notes_type = models.CharField(max_length=500)
class interviewtable(models.Model):
app_id = models.IntegerField(max_length=200)
interview = models.CharField(max_length=200)
interview_on = models.CharField(max_length=200)
interview_notes = models.CharField(max_length=200)
Upvotes: 1
Views: 3153
Reputation: 2836
Like this:
class interviewtable(models.Model):
app = models.ForeignKey(application)
interview = models.CharField(max_length=200)
interview_on = models.CharField(max_length=200)
interview_notes = models.CharField(max_length=200)
Django with automatically add id
thus app would be app_id
.
Also, you don't want to use max_length on integer field. If you want big integer use BigIntegerField()
Read the documentation properly: https://docs.djangoproject.com/en/1.5/
Upvotes: 4