Question

我正在尝试构建一个查询，该查询检索日期时间范围内的所有预订，并加入付款表以计算总付款，以及其他一些事情。该查询似乎工作得很好，除非它省略了没有付款的预订。任何帮助将不胜感激。

SELECT
    reservations.reservation_id ,
    properties. NAME ,
    properties.address1 AS prop_address ,
    reservations.last_name ,
    reservations.arrival_date ,
    reservations.departure_date ,
    reservations.date_created ,
    reservations.contract_filename ,
    reservations.contract_signed ,
    reservation_payments.date_of_payment ,
    (
        SUM(
            reservations.cleaning_fee + reservations.processing_fee + reservations.pool_heat_fee + reservations.special_fees + reservations.hoa_fees + reservations.rental_fee + reservations.taxes
        ) - COALESCE(
            SUM(
                reservation_payments.amount_paid
            ) ,
            0
        )
    ) AS sub_total ,
    COALESCE(
        SUM(
            reservation_payments.amount_paid
        ) ,
        0
    ) AS total_paid_to_date ,
    reservations.balance_due_date ,
    (
        SELECT
            GROUP_CONCAT(
                DISTINCT user_meta.first_name
                ORDER BY
                    associated_sales_staff.display_order ASC SEPARATOR ", "
            )
        FROM
            associated_sales_staff
        JOIN user_meta ON user_meta.user_id = associated_sales_staff.user_id
        WHERE
            (
                associated_sales_staff.reservation_id = reservations.reservation_id
            )
    ) AS sales_agents
FROM
    reservations
JOIN properties ON properties.prop_id = reservations.property_id
JOIN reservation_payments ON reservation_payments.reservation_id = reservations.reservation_id
WHERE
    properties.active_for_cleaning_schedule = 1
AND reservations.arrival_date >= "2017-05-15"
AND reservations.departure_date <= "2017-05-30"
GROUP BY
    reservations.reservation_id
HAVING
    (sub_total >= 1)
ORDER BY
    total_paid_to_date ASC

Answer 1

我发现问题出在reservation_payments表的JOIN上。

我更新了From：

JOIN reservation_payments ON reservation_payments.reservation_id = reservations.reservation_id

要：

LEFT JOIN reservation_payments ON reservations.reservation_id = reservation_payments.reservation_id

COALESCE（SUM（），0）JOINED表格列，GROUP BY不显示所有结果

1 个答案: