php - 从交叉引用表中获取唯一值

Question

现在我有一个交叉引用表，如下所示:

       Table job2pos
╔═══════════╦═════════════╗
║  job_id   ║ position_id ║
╠═══════════╬═════════════╣
║  1        ║  10         ║
║  2        ║  10         ║
║  2        ║  12         ║
║  3        ║  11         ║
║  3        ║  13         ║
║  4        ║  10         ║
║  5        ║  13         ║
╚═══════════╩═════════════╝

我想选择所有job_id有 postion_id = 10和position_id = 12 ，这将导致 job_id = 1,2,2,4 。但我只想要job_id = 2在我获取的结果中仅出现一次。

我尝试过的代码如下:

$positions = ["10","12"];

$sql = "SELECT job2pos.job_id FROM job2pos
WHERE job2pos.pos_id IN (?,?)";

$stmt = $pdo->prepare($sql);

foreach ($positions as $key => &$val) {
  $stmt->bindParam($key, $val, \PDO::PARAM_INT);
}

$stmt->execute();

var_dump($stmt->fetchAll(\PDO::FETCH_UNIQUE));

上述尝试将产生以下结果:

array(3) {
  [1]=> array(0) {}
  [2]=> array(0) {}
  [4]=> array(0) {}
}

所以我确实得到了唯一的值，并且额外的 2 不重复，但是我如何才能做到 job_id不作为索引出现，而是作为数组中的值出现？

• 附注如果您能指出我的代码中任何不必要的代码，那就太好了。

Answer 1

特殊的SELECT DISTINCT构造将仅从您的查询中带来不同的值。所以你可以将你的代码重写为:

$postions = ["10","12"];
$sql = "SELECT DISTINCT job2pos.job_id FROM job2pos WHERE job2pos.pos_id IN (?,?)";
$stmt = $pdo->prepare($sql);

// btw you don't have to bind elements one by one
// you can pass array as an argument to execute:
$stmt->execute($positions);

print_r($stmt->fetchAll(\PDO::FETCH_ASSOC));