php – Zend Framework 2 TableGateway返回空结果集

在最简单的表达式中,fetchAll函数有效：

public function fetchAll()
{
    $resultSet = $this->tableGateway->select();
    return $resultSet;
}

但是,当我尝试以下列方式混合连接时：

public function fetchAll()
{
    $sql = new Sql($this->tableGateway->getAdapter());

    $select = $sql->select();
    $select->from('Entreprise')
        ->columns(array('id','nom','categorie_id'))
        ->join(array('C' => 'Categorie'),'categorie_id = C.id',array('categorie' => 'nom'),ZendDbSqlSelect::JOIN_INNER);
    $resultSet = $this->tableGateway->selectWith($select);
    return $resultSet;
}

生成的查询是：

SELECT "Entreprise"."id" AS "id","Entreprise"."nom" AS "nom","Entreprise"."categorie_id" AS "categorie_id","C"."nom" AS "categorie" FROM "Entreprise" INNER JOIN "Categorie" AS "C" ON "categorie_id" = "C"."id"

表名,列都很好,因为查询不会抛出错误,而只是返回一个空结果集.即使删除连接,只是留下以下代码也无济于事.

public function fetchAll()
{
    $sql = new Sql($this->tableGateway->getAdapter());

    $select = $sql->select();
    $select->from('Entreprise');
    $resultSet = $this->tableGateway->selectWith($select);
    return $resultSet;
}

这让我相信我如何获得适配器或实例化选择是错误的,但我似乎无法弄明白或找到任何解决方案.

谁能帮我弄清楚我做错了什么？