Question

如果距离小于半径，我基本上想要返回所有用途。但是，如果连接到用户的元数据不存在，则不会返回结果。

继承我的疑问：

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( {$usersLatitude} ) ) * sin( radians( address.latitude ) ) ) ) AS distance
    FROM
        `users`
    JOIN
        `storage_varchars` AS `gender`
    ON
        gender.user_id = users.id AND gender.meta_name = 'gender'
    JOIN
        `storage_varchars` AS `sexual_orientation`
    ON
        sexual_orientation.user_id = users.id AND sexual_orientation.meta_name = 'sexual_orientation'
    JOIN
        `storage_varchars` AS `relationship_status`
    ON
        relationship_status.user_id = users.id AND relationship_status.meta_name = 'relationship_status'
    JOIN
        `storage_varchars` AS `interest_1`
    ON
        interest_1.user_id = users.id AND interest_1.meta_name = 'interest_1'
    JOIN
        `storage_varchars` AS `interest_2`
    ON
        interest_2.user_id = users.id AND interest_2.meta_name = 'interest_2'
    JOIN
        `storage_varchars` AS `interest_3`
    ON
        interest_3.user_id = users.id AND interest_3.meta_name = 'interest_3'
    JOIN
        `storage_varchars` AS `interest_4`
    ON
        interest_4.user_id = users.id AND interest_4.meta_name = 'interest_4'
    JOIN
        `storage_varchars` AS `interest_5`
    ON
        interest_5.user_id = users.id AND interest_5.meta_name = 'interest_5'
    JOIN
        `storage_varchars` AS `interest_6`
    ON
        interest_6.user_id = users.id AND interest_6.meta_name = 'interest_6'
    JOIN
        `payments` AS `address`
    ON
        address.user_id = users.id
) AS `main`
WHERE
    `main`.distance < {$radius}
ORDER BY
    `main`.distance

提前感谢。

Answer 1

以下是您的需求：

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( {$usersLatitude} ) ) * sin( radians( address.latitude ) ) ) ) AS distance
    FROM
        users
    LEFT JOIN
        storage_varchars AS gender
    ON
        users.id = gender.user_id AND gender.meta_name = 'gender'
    LEFT JOIN
        storage_varchars AS sexual_orientation
    ON
        users.id = sexual_orientation.user_id AND sexual_orientation.meta_name = 'sexual_orientation'
    LEFT JOIN
        storage_varchars AS relationship_status
    ON
        users.id = relationship_status.user_id AND relationship_status.meta_name = 'relationship_status'
    LEFT JOIN
        storage_varchars AS interest_1
    ON
        users.id = interest_1.user_id AND interest_1.meta_name = 'interest_1'
    LEFT JOIN
        storage_varchars AS interest_2
    ON
        users.id = interest_2.user_id AND interest_2.meta_name = 'interest_2'
    LEFT JOIN
        storage_varchars AS interest_3
    ON
        users.id = interest_3.user_id AND interest_3.meta_name = 'interest_3'
    LEFT JOIN
        storage_varchars AS interest_4
    ON
        users.id = interest_4.user_id AND interest_4.meta_name = 'interest_4'
    LEFT JOIN
        storage_varchars AS interest_5
    ON
        users.id = interest_5.user_id AND interest_5.meta_name = 'interest_5'
    LEFT JOIN
        storage_varchars AS interest_6
    ON
        users.id = interest_6.user_id AND interest_6.meta_name = 'interest_6'
    LEFT JOIN
        payments AS address
    ON
        users.id = address.user_id
) AS main
WHERE
    main.distance < {$radius}
ORDER BY
    main.distance;

SQL - 如果连接不存在则返回表

1 个答案: