SQL 元联接加上 Haversine 公式答案

【问题标题】：SQL meta joins plus Haversine formularSQL 元联接加上 Haversine 公式
【发布时间】：2014-01-14 22:22:01
【问题描述】：

我有一个查询，它将一些元数据连接到用户。

            SELECT
                users.*,
                gender.meta_value AS `gender`,
                sexual_orientation.meta_value AS `sexual_orientation`,
                relationship_status.meta_value AS `relationship_status`,
                interest_1.meta_value AS `interest_1`,
                interest_2.meta_value AS `interest_2`,
                interest_3.meta_value AS `interest_3`,
                interest_4.meta_value AS `interest_4`,
                interest_5.meta_value AS `interest_5`,
                interest_6.meta_value AS `interest_6`,
                address.address_line_1,
                address.address_line_2,
                address.town,
                address.county,
                address.postcode,
                address.country,
                address.longitude,
                address.latitude
            FROM
                `users`
            JOIN
                `storage_varchars` AS `gender`
            ON
                gender.user_id = users.id AND gender.meta_name = 'gender'
            JOIN
                `storage_varchars` AS `sexual_orientation`
            ON
                sexual_orientation.user_id = users.id AND sexual_orientation.meta_name = 'sexual_orientation'
            JOIN
                `storage_varchars` AS `relationship_status`
            ON
                relationship_status.user_id = users.id AND relationship_status.meta_name = 'relationship_status'
            JOIN
                `storage_varchars` AS `interest_1`
            ON
                interest_1.user_id = users.id AND interest_1.meta_name = 'interest_1'
            JOIN
                `storage_varchars` AS `interest_2`
            ON
                interest_2.user_id = users.id AND interest_2.meta_name = 'interest_2'
            JOIN
                `storage_varchars` AS `interest_3`
            ON
                interest_3.user_id = users.id AND interest_3.meta_name = 'interest_3'
            JOIN
                `storage_varchars` AS `interest_4`
            ON
                interest_4.user_id = users.id AND interest_4.meta_name = 'interest_4'
            JOIN
                `storage_varchars` AS `interest_5`
            ON
                interest_5.user_id = users.id AND interest_5.meta_name = 'interest_5'
            JOIN
                `storage_varchars` AS `interest_6`
            ON
                interest_6.user_id = users.id AND interest_6.meta_name = 'interest_6'
            JOIN
                `payments` AS `address`
            ON
                address.user_id = users.id

我现在想用它来使用 Haversine 公式通过最近距离进行搜索。

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20;

如何合并为两个？

问候

【问题讨论】：

标签： mysql haversine

【解决方案1】：

好的，我想出了一个方法，但谁能告诉我这是否正确？

$usersLatitude = '53.1765254';
$usersLongitude = '-1.1954137';
$distance = 20;

$query = "
    SELECT main.* FROM
    (
        SELECT
            users.*,
            gender.meta_value AS `gender`,
            sexual_orientation.meta_value AS `sexual_orientation`,
            relationship_status.meta_value AS `relationship_status`,
            interest_1.meta_value AS `interest_1`,
            interest_2.meta_value AS `interest_2`,
            interest_3.meta_value AS `interest_3`,
            interest_4.meta_value AS `interest_4`,
            interest_5.meta_value AS `interest_5`,
            interest_6.meta_value AS `interest_6`,
            address.address_line_1,
            address.address_line_2,
            address.town,
            address.county,
            address.postcode,
            address.country,
            address.longitude,
            address.latitude,
            ( 3959 * acos( cos( radians( {$usersLatitude} ) ) * cos( radians( address.latitude ) ) * cos( radians( address.longitude ) - radians( {$usersLongitude} ) ) + sin( radians( {$usersLatitude} ) ) * sin( radians( address.latitude ) ) ) ) AS distance
        FROM
            `users`
        JOIN
            `storage_varchars` AS `gender`
        ON
            gender.user_id = users.id AND gender.meta_name = 'gender'
        JOIN
            `storage_varchars` AS `sexual_orientation`
        ON
            sexual_orientation.user_id = users.id AND sexual_orientation.meta_name = 'sexual_orientation'
        JOIN
            `storage_varchars` AS `relationship_status`
        ON
            relationship_status.user_id = users.id AND relationship_status.meta_name = 'relationship_status'
        JOIN
            `storage_varchars` AS `interest_1`
        ON
            interest_1.user_id = users.id AND interest_1.meta_name = 'interest_1'
        JOIN
            `storage_varchars` AS `interest_2`
        ON
            interest_2.user_id = users.id AND interest_2.meta_name = 'interest_2'
        JOIN
            `storage_varchars` AS `interest_3`
        ON
            interest_3.user_id = users.id AND interest_3.meta_name = 'interest_3'
        JOIN
            `storage_varchars` AS `interest_4`
        ON
            interest_4.user_id = users.id AND interest_4.meta_name = 'interest_4'
        JOIN
            `storage_varchars` AS `interest_5`
        ON
            interest_5.user_id = users.id AND interest_5.meta_name = 'interest_5'
        JOIN
            `storage_varchars` AS `interest_6`
        ON
            interest_6.user_id = users.id AND interest_6.meta_name = 'interest_6'
        JOIN
            `payments` AS `address`
        ON
            address.user_id = users.id
    ) AS `main`
    WHERE
        `main`.distance < {$distance}
    ORDER BY
        `main`.distance
";

【讨论】：