如何在 ZF2 的 sql 语句中准备 sql 语句？

Question

我是 ZF2 新手，遇到以下问题：

在一个 db-table 中，我得到了两个引用另一个表的 id。如果我 选择他们，我想显示球员的名字，而不是 id 保存在表格中。

player (id, name)

board (id, playerA, playerB)

playerA 和 playerB 是整数，保存玩家表中的玩家 id。我已经与骨架应用程序相处融洽，但我可以传递给tableGateway->select() 的唯一子句是 where 子句。

有人有想法吗？如何在 ZF2 中准备这样的语句？

SELECT id, (SELECT name FROM player WHERE id = playerA) AS playerA, (SELECT name FROM player WHERE id = playerB) AS playerB FROM board

我找到的教程和参考资料都以 JOIN 结尾。但这不适合这里，对吧？

编辑： 以下是它，骨架应用程序可以正常工作

protected $select;

public function fetchAll()  {

$this->select = new Select();

$this->select->from('board')
    ->columns(array('id', 'kindOf', ...))
    ->join(['pA' => 'player'], 'pA.id = board.playerA', ['playerA' => 'name'])
    ->join(['pB' => 'player'], 'pB.id = board.playerB', ['playerB' => 'name']);

$resultSet = $this->tableGateway->selectWith($this->select);
return $resultSet;
}

Answer 1

您的问题不是 ZF2 问题，而是 SQL 问题。例如，您可以这样写：

SELECT id, pA.name AS playerA, pB.name AS playerB
FROM board
INNER JOIN player AS pA ON pA.id=board.playerA
INNER JOIN player AS pB ON pB.id=board.playerB

在 ZF2 中：

$select = new Select();
$select->from('board')
       ->columns('id')
       ->join(['pA' => 'player', 'pA.id = board.playerA', ['playerA' => 'name'])
       ->join(['pB' => 'player', 'pB.id = board.playerB', ['playerB' => 'name'])