Question

查询：

SELECT
    c.id campaign_id,
    c.campaign_title,
    c.start_date campaign_start_date,
    c.end_date campaign_end_date,
    cat.id category_id,
    cat.category_title,
    n.id nominee_id,
    n.title nominee_title,
    u.id voter_id,
    u.fullname voter_name
FROM
    campaign c
LEFT JOIN categories cat ON cat.campaign_id = c.id
LEFT JOIN category_nominees cn ON cn.category_id = cat.id
LEFT JOIN nominees n ON n.id = cn.nominee_id
LEFT JOIN category_votes cv ON cv.campaign_id = c.id
AND cv.category_id = cat.id
AND cv.nominee_id = n.id
LEFT JOIN users u on u.id = cv.user_id
WHERE
    c.active = 1
ORDER BY
    u.fullname,
    c.campaign_title,
    cat.category_order,
    cat.category_title,
    cn.nominee_order,
    n.title

使用该查询，我将使用PHP在嵌套的关联数组中逻辑存储所有数据，然后无论何时循环数据或需要引用数据，您都会引用数组指针，而不是每次都执行新的查询。

例如： $ campaign [$ campaign_id] ['categories'] [$ category_id] ['nominees'] [$ nominee_id] ['title']是一种可能的方法。

我试过

$rows = [];
while($row = mysqli_fetch_array($query))
{
    if (!isset($rows[$row['campaign_id']])) {
        $rows[$row['campaign_id']] = array(
            'campaign_id' => $row['campaign_id'],
            'category_id' => array()
        );
    }

    array_push ($rows[$row['campaign_id']]['category_id'],array(
        'category_id' => $row['category_title'],
        'nominees_id' => $row['nominee_id']
    ));

}

我无法理解。任何帮助将不胜感激！

Answer 1

$campaigns = array(); while($row = mysqli_fetch_array($query)) {
    $campaign_id = $row['campaign_id'];
    $category_id = $row['category_id'];
    $nominee_id = $row['nominee_id'];
    $campaigns[$campaign_id]['categories'][$category_id]['nominees'][$nominee_id]['title']
= $row; }

如何从此查询创建多维数组

1 个答案: