我刚刚开始使用Zend Framework 2进行开发,我遇到了障碍.
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'), \Zend\Db\Sql\Select::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?
最佳答案 以下代码完美运行:
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'), \Zend\Db\Sql\Select::JOIN_INNER);
$resultSet = $this->tableGateway->selectWith($select);
return $resultSet;
}
…这是在另一个文件中意外输入的错字…
感谢Sam的领导,可笑的简单,但我只是没想到尝试!