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ターミナルに送り返すためにJSONに入れる必要があるやや複雑なphp sqlクエリがあります。誰か助けてくれませんか？私はこれに苦労しており、正しく理解できないようです。私のコードは次のようなものです：

//Grab all people living in a city, and for each of those people, grab all of their carsmodel and license plate.  My output should be something like [{"firstname":John, "lastname":Smith, "cars":[car1, car2, car3...], "plates":[plate1, plate2, ...]},{"firstname":Jack,"lastname":Hide, "cars":[car1, car2, car3], "plates":[plate1, plate2,...]}]
$sql = "SELECT id, firstname, lastname FROM residents WHERE residents.city = ?";
$q = $connection->prepare($sql);
$q->execute(array($city));

while($row = $q->fetch())
{
    $sql2 = "SELECT carid FROM cars WHERE userid = ?"
    $q2 = $connection->prepare($sql2);
    $q2->execute(array($row[0]));
    while($row2 = $q2->fetch())
    {
        // What do I do here?!
    }
}

return json_encode(//?);

どんな助けでも大歓迎です！

ありがとう！

4

2 に答える 2

1

1 つのクエリで作成します。

SELECT id, firstname, lastname, carid -- and perhaps other cars columns here
FROM residents
INNER JOIN cars
ON cars.userid = residents.id
WHERE residents.city = ?
ORDER BY residents.id

次にPHP部分で：

$data = array();
$current = array();
while($row2 = $q2->fetch())
{
    if ($current['id'] != $row[0]) {
        $data[] = $current;          
        $current = array('id' => $row[0], 'firstname' => $row[1], 'lastname' => $row[2] );
    } else {
        $current['cars'][] = array( /* row data for car */ );
    }
}
array_shift($data);
return json_encode($data);
于 2013-03-29T22:20:54.680 に答える
1

数値を操作する代わりに、フェッチ モードをPDO::FETCH_OBJ(または) に設定します。PDO::FETCH_ASSOC接続でこれを行うことができますが、fetch.

$q2 = $connection->prepare("SELECT carid FROM cars WHERE userid = ?");
$users = array();
while ($row = $q->fetch(PDO::FETCH_OBJ)) {
    $q2->execute(array($row->id));
    $row->cars = $q2->fetchall(PDO::FETCH_OBJ);
    $users[] = $row;
}
return json_encode($users);
于 2013-03-29T22:07:29.817 に答える