missing FROM-clause entry for table “Grupo” cakephp

Question

hi i have aproblem in my code, I want generate a list of user but this have a group and need just a group of user. the error say:

Error: SQLSTATE[42P

Answer 1

You need the grupos table to be joined in the query, your query in the question has no joins. There are a number of simple solutions.

Define recursive.

Recursive is a very coarse control of what joins and queries are executed, by default find('list') has a recursive value of -1.

-1 means no joins, which is why there is no join in the resultant query. Setting it to a value of 0 adds a join to the main query for all hasOne and belongsTo associations.

Be wary of using/relying on recursive as it's very easy to generate queries with joins you don't need - and/or triggering many subsequent queries for related data (if set to a value larger than 0).

However this find call:

$data = $this->Soya->find('list', array(
    'fields'=> array('Soya.id','Soya.username'),
    'recursive' => 0, // added
    'conditions' => array(
        'Grupo.categoria' => 'Soya' , 
        'Grupo.subcategoria' => 'Productor de Oleaginosas'
    )
));

Should result in this query (If the Soya model has a belongsTo association to Grupo):

SELECT
    "Soya"."id" AS "Soya__id",
    "Soya"."username" AS "Soya__username"
FROM
    "public"."users" as "Soya"
LEFT JOIN
    "public"."Grupos" as "Grupo" on ("Soya"."grupo_id" = "Grupo"."id")
...
Possibly more joins
...
WHERE
   "Grupo"."categoria" = 'Soya' 
    AND 
    "Grupo"."subcategoria" = 'Productor de Oleaginosas'

Or Use containable

The containable behavior allows better control of what queries are executed. Given the info in the question to use it that means:

Will permit you to do the following in your controller:

$data = $this->Soya->find('list', array(
    'fields'=> array('Soya.id','Soya.username'),
    'contain' => array('Grupo'), // added
    'conditions' => array(
        'Grupo.categoria' => 'Soya' , 
        'Grupo.subcategoria' => 'Productor de Oleaginosas'
    )
));

Which will generate the following query (exactly one join):

SELECT
    "Soya"."id" AS "Soya__id",
    "Soya"."username" AS "Soya__username"
FROM
    "public"."users" as "Soya"
LEFT JOIN
    "public"."Grupos" as "Grupo" on ("Soya"."grupo_id" = "Grupo"."id")
WHERE
   "Grupo"."categoria" = 'Soya' 
    AND 
    "Grupo"."subcategoria" = 'Productor de Oleaginosas'