5

我希望有人可以帮助我解决这个问题。

假设我有 3 个数据库表：

Users:
user_id, user_name
100, John
101, Jessica

Cars:
car_id, car_name
30, Corvette
31, BMW


UsersCars:
user_id, car_id, car_colour
100, 30, Red
101, 30, Green
101, 31, Green

(so John got a red corvette and Jessica has a green Corvette and a BMW)

我想要返回一个多维 PHP 数组的代码，例如：

Array
(
    [100] => Array
    (
        [user_id] => 100
        [user_name] => John
        [cars] => Array
        (
            [car_id]=>30,
            [car_name]=>'Corvette',
            [car_colour]=>'Red'

        )            
    )
    [101] => Array
    (
        [user_id] => 101
        [user_name] => Jessica
        [cars] => Array
        (
            [0] => Array
            (
                [car_id]=>30,
                [car_name]=>'Corvette',
                [car_colour]=>'Green'
            ),
            [1] => Array
            (
                [car_id]=>31,
                [car_name]=>'BMW',
                [car_colour]=>'Green'
            )
        )            
    )
)

我有以下 SQL

SELECT u.*, c.* FROM Users u
LEFT JOIN UsersCars uc ON u.user_id = uc.user_id
LEFT JOIN Cars c ON uc.car_id = c.car_id

和 PHP

$result = mysqli_query($db, $q);
while ($row = mysqli_fetch_assoc($result)) {
    $users_with_cars[$row['user_id']] = $row;
}

但这是不正确的。任何人都知道如何解决导致上述数组的问题（考虑到性能）？我不想硬编码“汽车”的例外情况，因为它可能会发生不止一次。我宁愿有一些只查看 $row 和 $users_with_cars 的东西，当看到一些新值时，它会通过将旧值转换为数组来附加它。也许已经有一个原生的 PHP 函数了？或者更好，也许我的 MySQL 或整个方法是错误的？

任何帮助或提示表示赞赏。

问候

更新已解决

这是一个更新，也许我可以帮助其他人我最终如何解决它。

我最终总是为一辆或多辆汽车使用一个数组，并且我调整了表格以始终将“id”作为列名。这样您就可以轻松扩展它。见例子；

Users:
id, name
100, John
101, Jessica

Cars:
id, name
30, Corvette
31, BMW


UsersCars:
user_id, car_id, car_colour
100, 30, Red
101, 30, Green
101, 31, Green



$q = 'SELECT u.*, c.id as car_id, c.name as car_name, uc.colour as car_colour FROM Users u
LEFT JOIN UsersCars uc ON u.id = uc.user_id
LEFT JOIN Cars c ON uc.car_id = c.id';

$result = mysqli_query($db, $q);
while ($row = mysqli_fetch_assoc($result)) {
    $users_with_cars[] = $row;
}
$joins = array('cars' => array('car_id'=>'id','car_name'=>'name','car_colour'=>'colour'));
$users_with_cars = create_join_array($users_with_cars, $joins);

print_r($users_with_cars);



function create_join_array($rows, $joins){
    /* build associative multidimensional array with joined tables from query rows */

    foreach((array)$rows as $row){
        if (!isset($out[$row['id']])) {
            $out[$row['id']] = $row;
        }

        foreach($joins as $name => $item){
            unset($newitem);
            foreach($item as $field => $newfield){
                unset($out[$row['id']][$field]);
                if (!empty($row[$field]))
                    $newitem[$newfield] = $row[$field];
            }
            if (!empty($newitem))
                $out[$row['id']][$name][$newitem[key($newitem)]] = $newitem;
        }
    }

    return $out;
}

这一切都产生了美丽的数组：

Array
(
    [100] => Array
    (
        [id] => 100
        [name] => John
        [cars] => Array
        (
            [30] => Array
            (
                [id]=>30
                [name]=>'Corvette',
                [colour]=>'Red'
            )
        )            
    )
    [101] => Array
    (
        [id] => 101
        [name] => Jessica
        [cars] => Array
        (
            [30] => Array
            (
                [id]=>30,
                [name]=>'Corvette',
                [colour]=>'Green'
            ),
            [31] => Array
            (
                [id]=>31,
                [name]=>'BMW',
                [colour]=>'Green'
            )
        )            
    )
)

假设用户也可以拥有多辆自行车。然后，您有多个连接数组，您可以轻松地使用左连接绑定并将其添加到连接数组。

$q = 'SELECT u.*, c.id as car_id, c.name as car_name, uc.colour as car_colour, b.id as bike_id, b.name as bike_name FROM Users u
LEFT JOIN UsersCars uc ON u.user_id = uc.user_id
LEFT JOIN Cars c ON uc.car_id = c.id
LEFT JOIN UsersBikes ub ON u.user_id = ub.user_id
LEFT JOIN Bikes b ON ub.bike_id = b.id';

$result = mysqli_query($db, $q);
while ($row = mysqli_fetch_assoc($result)) {
    $users_with_cars_bikes[] = $row;
}

$joins = array('cars' => array('car_id'=>'id', 'car_name'=>'name', 'car_colour'=>'colour'), 
               'bikes' => array('bike_id'=>'id', 'bike_name'=>'name'));
$users_with_cars_bikes = create_join_array($users_with_cars_bikes, $joins);

print_r($users_with_cars_bikes);

会导致类似

Array(
    [100] => Array
    (
        [id] => 100
        [name] => John
        [cars] => Array
        (
            [30] => Array
            (
                [id]=>30
                [name]=>'Corvette',
                [colour]=>'Red'
            )
        )
        [bikes] => Array
        (
            [41] => Array
            (
                [id]=>41
                [name]=>'BMX'
            )
        )           
    )
)

等等..

谢谢大家帮忙:)

4

2 回答 2

2

这是我能想到的。它将（可能）创建一个数组，就像您想要的输出一样。不确定它是否会起作用，我在这里写了这个没有测试。让我知道！:)

$result = mysqli_query($db, $q);
while ($row = mysqli_fetch_assoc($result)) 
{
    // Add user ID and name to the array
    $users_with_cars[$row['user_id']]['user_id'] = $row['user_id'];
    $users_with_cars[$row['user_id']]['user_name'] = $row['user_name'];
    // Check if this user has cars in the array. If not, this is the first car (see else)
    if(isset($users_with_cars[$row['user_id']]['cars']))
    {
        // Check if there is exactly 1 car in the array
        if(count($users_with_cars[$row['user_id']]['cars']) == 1)
        {
            // If yes, put that car in a 'sub array'
            $users_with_cars[$row['user_id']]['cars'] = array(0 => $users_with_cars[$row['user_id']]['cars']);
            // Then add the new car
            $users_with_cars[$row['user_id']]['cars'][] = array('car_id' => $row['car_id'], 'car_name' => $row['car_name'], 'car_color' => $row['car_color']);
        }
        else
        {
            // It already has more than one car in the array. Just add it
            $users_with_cars[$row['user_id']]['cars'][] = array('car_id' => $row['car_id'], 'car_name' => $row['car_name'], 'car_color' => $row['car_color']);
        }
    }
    else
    {
        // Add a single car without 'sub array'
        $users_with_cars[$row['user_id']]['cars']['car_id'] = $row['car_id'];
        $users_with_cars[$row['user_id']]['cars']['car_name'] = $row['car_name'];
        $users_with_cars[$row['user_id']]['cars']['car_color'] = $row['car_color'];
    }
}

编辑：

这是我在评论中提到的一个例子：

$result = mysqli_query($db, $q);
while ($row = mysqli_fetch_assoc($result)) 
{
    $count = count($users_with_cars[$row['user_id']]['cars']);
    foreach($row AS $key => $value)
    {
        if(substr($key, 0, 4) == "car_")
        {
            // Single:
            $users_with_cars[$row['user_id']]['cars'][$key] = $value;
            // Multiple:
            $users_with_cars[$row['user_id']]['cars'][$count][$key] = $value;
        }
    }
}
于 2012-11-09T15:29:01.890 回答
0 
$result = mysqli_query($db, $q);
$users_with_cars = array();
while ($row = mysqli_fetch_assoc($result)) {
    if (!isset($users_with_cars[$row['user_id']])) {
        $user = array();
        $user['user_id'] = $row['user_id'];
        $user['user_name'] = $row['user_name'];
        $user['cars'] = array();
        $users_with_cars[$row['user_id']] = $user;
    }
    $car = array();
    $car['car_id'] = $row['car_id'];
    $car['car_name'] = $row['car_name'];
    $car['car_colour'] = $row['car_colour'];
    $users_with_cars[$row['user_id']]['cars'] = $car;
}
于 2012-11-09T14:56:26.410 回答