rewriting a getresource query into a snippet

17 Posts

dominikwilkowski Reply #11, 13 years, 3 months ago

No returns for either copy past your code or this:

$query->leftJoin('modTemplateVarResource', 'tv13', 'tv13.contentid = modResource.id'); //with a custom join-alias, you need to define a ON - statement
$query->where(array(
   'modResource.parent' => 679,
   'modResource.published' => 1,
   'tv13.tmplvarid' => 13,
   'tv13.value' => 603
));

:(

5,888 Posts

Bruno17 Reply #12, 13 years, 3 months ago

Quote from: Bruno17 at Jul 12, 2013, 03:40 AM

what do you get with
$query->prepare();echo $query->toSql();
try to see what you get with this generated query in phpmyadmin

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17 Posts

dominikwilkowski Reply #13, 13 years, 3 months ago

That returns:

ERROR 1054 (42S22): Unknown column 'tv_value.resourceid' in 'on clause'

I played around with that but I'm really taping in the dark here.
Thanks for the help!

17 Posts

dominikwilkowski Reply #14, 13 years, 3 months ago

I got it!

The query is:

$query->leftJoin('modTemplateVarResource', 'tv_value', 'tv_value.contentid = modResource.id');

I looked into the query and database and now understand what that translates to.
The

$query->prepare();echo $query->toSql();

really helped here!

Thanks @bruno17