zend-framework2 - ZF2 和嵌套集 - 如何使用 Select 对象进行交叉连接？

Question

要在嵌套集中获取树，我们必须做一个笛卡尔积（每个与每个）。

select * from tree as t1, tree as t2

或者

select * from tree as t1 cross join tree as t2

使用 zf1 很容易，但是在 zf2 中使用\Zend\Db\Sql\Select对象时没有交叉连接？

有没有办法使用 Select() 来实现它？还是纯 sql 查询是唯一的方法？

不幸的是，谷歌搜索失败了，所以提前感谢您的任何建议。

Answer 1

可以将连接类型作为最后一个参数传递

class Select extends AbstractSql implements SqlInterface, PreparableSqlInterface
{
    const JOIN_INNER = 'inner';
    const JOIN_OUTER = 'outer';
    const JOIN_LEFT = 'left';
    const JOIN_RIGHT = 'right';
    const SQL_STAR = '*';
    const ORDER_ASCENDING = 'ASC';
    const ORDER_DESENDING = 'DESC';

    public $where; // @param Where $where

    public function __construct($table = null);
    public function from($table);
    public function columns(array $columns, $prefixColumnsWithTable = true);
    public function join($name, $on, $columns = self::SQL_STAR, $type = self::JOIN_INNER);
    public function where($predicate, $combination = Predicate\PredicateSet::OP_AND);
    public function group($group);
    public function having($predicate, $combination = Predicate\PredicateSet::OP_AND);
    public function order($order);
    public function limit($limit);
    public function offset($offset);
}

所以，zf1 方法joinCross(table, [columns])等价于 zf2join($name, $on, $columns, 'cross')

Answer 2

CROSS JOIN 操作可以替换为 (INNER) JOIN，其中 join 子句的计算结果始终为 true (1=1)

select * from tree as t1 inner join tree as t2 on 1=1

Zf2 示例：

$select = new Sql\Select;

$select ->from(array('node' => 'nested_category'))
        ->columns(array('node_name' => new Sql\Expression('CONCAT(REPEAT(" ", COUNT(parent.name)-1), node.name)')))
        ->join(array('parent' => 'nested_category'), new Sql\Expression('1=1'), array('id'))
        ->where('node.lft BETWEEN parent.lft AND parent.rgt')
        ->group('node.name')
        ->order('node.lft');

$resultSet = $this->selectWith($select);