Question

我正在构建一个SQL查询 1）创建一个group_concat，如果值不为null，则返回json 2）将一个表（句柄）连接到两个不同的表。

我在查询时遇到两个问题 1）当我得到'应该'失败并且返回0行的结果时，我得到一行，其中所有字段都为空 2）当我有一个没有活动的scene时，我最终得到一个JSON_OBJECT，其中所有字段都为空

    SELECT h.name as name, s.*, 
        GROUP_CONCAT(JSON_OBJECT(
            'activityId', a.id,
            'name', ah.name,
            'settings', a.settings,
        ) SEPARATOR ',') as activities
        FROM scene AS s
        INNER JOIN user AS h ON s.uid = h.id
        LEFT JOIN scene_activities AS sa ON sa.sid = s.id
        LEFT JOIN activity AS a ON sa.aid = a.id
        LEFT JOIN user AS ah ON a.uid = ah.id
  WHERE s.scene_id = ?

如何设置上述字段，以便不返回Scene的空集或JSON_OBJECT的空集为空。

Answer 1

如果您要求json中的所有属性都具有非空值，请不要使用left连接（如果没有匹配的行，则返回空值）并在其上添加条件连接行中的值也是：

SELECT h.name as name, s.*, 
    GROUP_CONCAT(JSON_OBJECT(
        'activityId', a.id,
        'name', ah.name,
        'settings', a.settings,
    ) SEPARATOR ',') as activities
FROM scene AS s
JOIN user AS h ON s.uid = h.id
JOIN scene_activities AS sa ON sa.sid = s.id
JOIN activity AS a ON sa.aid = a.id
    and a.settings is not null
JOIN user AS ah ON a.uid = ah.id
    and ah.name is not null
WHERE s.scene_id = ?

mysql 4 way join为所有字段返回null

1 个答案: