Zend DB Selecting constants - columns that do not exist in table

Question

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

score 20 · Accepted Answer · answered Jul 22 '10 at 06:24

20

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'));

answered Jul 22 '10 at 06:24

robertbasic

4,325
1
27
25

Worked for me too. Thanks – Ehsan Jul 17 '17 at 01:08

score 2 · Answer 2 · answered May 19 '11 at 21:44

2

$select->from(
    array('sh'=>'shopping'),
    array('shopping_id','shopping_details','friend'=>'type', 'alias'=>'column or expression')
);

answered May 19 '11 at 21:44

Ballsacian1

17,214
2
26
25

score 0 · Answer 3 · answered Nov 02 '21 at 13:40

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.

Zend DB Selecting constants - columns that do not exist in table

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