In my search method I have the following code …
$query->joinWith(["userFrom", "userTo"]);
This is 2 foreign keys that link to the same user table.
This generates the following error …
Syntax error or access violation: 1066 Not unique table/alias: ‘user’
Can anyone tell me how to get around this?
In your model you need to define two aliases for the same table.
Like for example:
public function getUserFrom()
{
return $this->hasOne(User::className(), ['id' => 'user_id'])
->from(User::tableName().' AS userFrom');
}
public function getUserTo()
{
return $this->hasOne(User::className(), ['id' => 'user_id'])
->from(User::tableName().' AS userTo');
}
Regards
Thanks, it works!
No Prob.
Glad I could help. 
The problem is I also have this code in a parent model …
public static function find() {
return parent::find()->where([static::tableName() . ".deleted" => 0]);
}
Which now generates the following error because the tableName is no longer correct …
Column not found: 1054 Unknown column ‘user.deleted’ in ‘where clause’
Because the user.deleted is no longer user.deleted but userFrom.deleted etc.
Is there anyway around this?
Sorry, I have not overwritten default find() method so far…
Means: I don’t know yet.
When I have time I will try to reproduce and check later
Some Questions:
Where do you receive the error?
Only in your ModelSearch? Or everywhere?
Do you have deleted getUser() method in model?
If yes: what happens when you add it back additional to getUserFrom(), getUserTo()?
Regards
I only receive the error when the search code is used.
No I have no getUser method in model only getUserFrom and getUserTo.
If I put the method back in, the error still appears.
u have to define with table name.
like
tbl_city.cityId
What do you mean, where do I define with table name?
Anyone got any ideas for this one.
The first bit works with the aliases for the relations but because I have override the find() method like this …
public static function find() {
return parent::find()->where([static::tableName() . ".deleted" => 0]);
}
But because now the tableName() will be different, because of the use of the alias it is throwing an error.