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Zend Db Select ? Substitution in join* condition

It doesn't look like there's any parameter substitution in Zend_Db_Select's on clause.

It's highly annoying that I can't just do something like:

$select->joinLeft('st_line_item','st_line_item.order_id = st_order.id and st_line_item.status = ?')

So what's the idiomatic alternative that works within the fluent interface? I could do something like prepare the join clause on the outside but that's not the point.

like image 593
Koobz Avatar asked Feb 02 '10 01:02

Koobz


2 Answers

This should work:

$select->joinLeft(
    'st_line_item',
    $this->_db->quoteInto(
        'st_line_item.order_id = st_order.id and st_line_item.status = ?', 
        $param
    )
)

Basically, anytime you want to escape a variable where a Zend_Db_* method doesn't do it automatically, you just use Zend_Db::quoteInto() to do the job.

like image 177
Noah Goodrich Avatar answered Nov 18 '22 16:11

Noah Goodrich


This is how I always do it, it's not a work of art but it gets the job done:

$param = $db->quote($param);
$select->joinLeft(
    'st_line_item',
    'st_line_item.order_id = st_order.id and st_line_item.status = ' . $param
);
like image 1
Mike B Avatar answered Nov 18 '22 17:11

Mike B