Question

我有3张桌子：

诊所

id | name | description | lat | lng | opening_hours| logo | address | city | zip | phone_number | email | url | gmaps_link | marker | created_at | updated_at

服务

id | name | created_at | updated_at

CLINIC_SERVICES

clinic_id | service_id

此查询应返回两个结果，但此刻返回0：

SELECT * FROM
        (SELECT clinics.id as cid, clinics.lat, clinics.lng, clinics.opening_hours,
        clinics.logo, clinics.address, clinics.city, clinics.name, clinics.description,
        clinics.zip, clinics.phone_number, clinics.email, clinics.url, clinics.gmaps_link,
        clinics.marker,
        countries.full_name AS country,
        (6378 * acos(
            cos(radians(-33.84801)) * cos(radians(lat)) *
            cos(radians(lng) - radians(151.06488)) +
            sin(radians(-33.84801)) * sin(radians(lat))))
        AS distance
        FROM clinics
        JOIN countries ON countries.id = clinics.country_id
        LEFT JOIN clinics_services ON clinics.id = clinics_services.clinic_id
         WHERE clinics_services.service_id = 1 AND clinics_services.service_id = 29
        GROUP BY clinics.id
        ) AS distances
    WHERE distance < 50000
    ORDER BY distance ASC

如果我输入 OR 而不是 AND ，我将获得5个诊所，这些诊所实际上可以正常工作。如何获得正确的结果（诊所同时提供两种服务）？我在这里做什么错了？

Answer 1

根据previous similar question，此方法应该有效：

SELECT * FROM
        (SELECT clinics.id as cid, clinics.lat, clinics.lng, clinics.opening_hours,
        clinics.logo, clinics.address, clinics.city, clinics.name, clinics.description,
        clinics.zip, clinics.phone_number, clinics.email, clinics.url, clinics.gmaps_link,
        clinics.marker,
        countries.full_name AS country,
        (6378 * acos(
            cos(radians(-33.84801)) * cos(radians(lat)) *
            cos(radians(lng) - radians(151.06488)) +
            sin(radians(-33.84801)) * sin(radians(lat))))
        AS distance
        FROM clinics
        JOIN countries ON countries.id = clinics.country_id
        LEFT JOIN clinics_services ON clinics.id = clinics_services.clinic_id
         WHERE clinics_services.service_id = 1 OR clinics_services.service_id = 29
        GROUP BY clinics.id
        HAVING SUM(clinics_services.service_id = 1) > 0 AND SUM(clinics_services.service_id = 29) > 0
        ) AS distances
    WHERE distance < 50000
    ORDER BY distance ASC

MySQL-带枢轴表的WHERE / AND

1 个答案: