问题
I have this (simplified) table structure:
users
- id
- type (institutions or agents)
institutions_profile
- id
- user_id
- name
agents_profile
- id
- user_id
- name
And I need to create a profile
relationship on the Users
model, but the following doesn't work:
class User extends Model
{
public function profile()
{
if ($this->$type === 'agents')
return $this->hasOne('AgentProfile');
else
return $this->hasOne('InstitutionProfile');
}
}
How could I achieve something like that?
回答1:
Lets take a different approach in solving your problem. First lets setup relationship for the various models respectively.
class User extends Model
{
public function agentProfile()
{
return $this->hasOne(AgentProfile::class);
}
public function institutionProfile()
{
return $this->hasOne(InstitutionProfile::class);
}
public function schoolProfile()
{
return $this->hasOne(SchoolProfile::class);
}
public function academyProfile()
{
return $this->hasOne(AcademyProfile::class);
}
// create scope to select the profile that you want
// you can even pass the type as a second argument to the
// scope if you want
public function scopeProfile($query)
{
return $query
->when($this->type === 'agents',function($q){
return $q->with('agentProfile');
})
->when($this->type === 'school',function($q){
return $q->with('schoolProfile');
})
->when($this->type === 'academy',function($q){
return $q->with('academyProfile');
},function($q){
return $q->with('institutionProfile');
});
}
}
Now you can access your profile like this
User::profile()->first();
This should give you the right profile. Hope it helps.
回答2:
you can do this by use another method please check this:
a blog Post and Video model could share a polymorphic relation to a Tag model. Using a many-to-many polymorphic relation allows you to have a single list of unique tags that are shared across blog posts and videos. First, let's examine the table structure:
https://laravel.com/docs/5.4/eloquent-relationships#many-to-many-polymorphic-relations
回答3:
Looks like that should be $this->type
rather than $this->$type
- since type
is a property, not a variable.
来源:https://stackoverflow.com/questions/43668153/how-to-setup-conditional-relationship-on-eloquent