Reputation: 2618
I'm trying to do this query using Zend DB select but I'm not able to do so
This is the sql query
select shopping_id,shopping_details,"friend" as type from shopping
Notice here how I'm specifying "friend" as type and friend is not a column in the shopping table.
Now how do I do this in Zend. I have tried this but it gives me an error saying "sh.friend Column does not exist"
$select->from(array('sh'=>'shopping'),array('shopping_id','shopping_details','"friend" as type');
Any help will be appreciated thanks
Upvotes: 11
Views: 5730
Reputation: 3804
For Zend Framework 2/3 or Laminas you have to use Laminas\Db\Sql\Expression
. Make sure to quote your constant with a double-quote ""
.
$select->from(['e' => 'experience'])
->columns([
'id' => 'id',
'value' => 'title',
'name' => new Laminas\Db\Sql\Expression('"skill"')
]);
*for Zend Framework the name of the expression class is Zend\Db\Sql\Expression
.
Upvotes: 0
Reputation: 17322
$select->from(
array('sh'=>'shopping'),
array('shopping_id','shopping_details','friend'=>'type', 'alias'=>'column or expression')
);
Upvotes: 2
Reputation: 4335
Try with Zend_Db_Expr
, maybe something like:
$select->from(array('sh'=>'shopping'),
array('shopping_id','shopping_details',
new Zend_Db_Expr('"friend" as type'));
Upvotes: 20