Question

我有一个带有多个子查询的复杂mysql查询。我需要其中一个子查询与主表相关。那是我的疑问：

SELECT table_cl.id, table_cl.nombre, table_cl.apellidos, (SELECT COUNT(*) FROM (
            SELECT iauno.id_accion, iauno.id_cliente, iauno.asistencia
            FROM inter_clientes_acciones iauno
            INNER JOIN clientes clientesuno ON clientesuno.id = iauno.id_cliente
            WHERE iauno.id_cliente IN(
                SELECT iasubuno.id_cliente
                FROM inter_clientes_acciones iasubuno
                INNER JOIN acciones acsubuno ON iasubuno.id_accion = acsubuno.id
                WHERE acsubuno.id_proyecto = 3
                GROUP BY iasubuno.id_accion 
                HAVING COUNT(*) = 1
            ) 
        )caa
    ) as convocado,
    (SELECT COUNT(*) FROM (
            SELECT iados.id_accion, iados.id_cliente, iados.asistencia
            FROM inter_clientes_acciones iados
            INNER JOIN clientes cldos ON cldos.id = iados.id_cliente
            WHERE iados.id_cliente IN(
                SELECT iasubdos.id_cliente
                FROM inter_clientes_acciones iasubdos
                INNER JOIN acciones acsubdos ON iasubdos.id_accion = acsubdos.id
                WHERE acsubdos.id_proyecto = 3 AND iasubdos.id_cliente = table_cl.id
                GROUP BY id_accion 
                HAVING COUNT(*) = 1
            ) 
        )car
    ) as realizado 
FROM clientes table_cl
WHERE table_cl.id_proyecto = 3 AND table_cl.id IN (
    SELECT iatres.id_cliente
    FROM inter_clientes_acciones iatres
    INNER JOIN clientes cltres ON cltres.id = iatres.id_cliente
    WHERE iatres.id_cliente IN(
        SELECT iasubtres.id_cliente
        FROM inter_clientes_acciones iasubtres
        INNER JOIN acciones acsubtres ON iasubtres.id_accion = acsubtres.id
        WHERE acsubtres.id_proyecto = 3
        GROUP BY iasubtres.id_accion 
        HAVING COUNT(*) = 1
        ) 
)

当我执行查询时，出现错误Error Code: 1054. Unknown column 'table_cl.id' in 'where clause'

我如何修改查询以引用table_cl上的id

修改

可能需要解释：我有3个表，客户端，操作和clients_actions（用西班牙语表示：clientes，acciones和inter_clientes_acciones）。可以为一个客户端或多个客户端创建一个操作。此查询的目的是获取以下字段：客户端的名称和姓氏，分配给客户端的操作数，以及客户端已协助的此操作的数量，但仅指定分配给他的操作（单个操作）。 / p>

这里有一个sql fiddle，其中包含架构和我所做的失败的SQL。

- ＆GT;编辑2： 我找到了另一种获取数据的方法。在这种情况下，我将字段中的所有操作ID分组，并且（在PHP文件中）我将其计算在内。我收到了这个回复：

Individual actions: [ { "nombre": "Susana", "apellidos": "Rodr\u00edguez Torr\u00f3n", "acc_convocadas": "2,3,5", "acc_realizadas": "2" }, { "nombre": "Pablo", "apellidos": "Campos P\u00e9rez", "acc_convocadas": "5", "acc_realizadas": "-" }, { "nombre": "Oscar", "apellidos": "Delacroix", "acc_convocadas": "-", "acc_realizadas": "-" } ]
Group actions: [ { "nombre": "Susana", "apellidos": "Rodr\u00edguez Torr\u00f3n", "acc_convocadas": "3,1,4", "acc_realizadas": "-" }, { "nombre": "Pablo", "apellidos": "Campos P\u00e9rez", "acc_convocadas": "-", "acc_realizadas": "-" }, { "nombre": "Oscar", "apellidos": "Delacroix", "acc_convocadas": "3", "acc_realizadas": "-" } ]

这是

条款

SELECT nombre, apellidos,
  IFNULL((SELECT GROUP_CONCAT(id_accion)
     FROM inter_clientes_acciones iac
     WHERE iac.id_cliente = cl.id AND id_accion IN (
       SELECT iauno.id_accion
       FROM inter_clientes_acciones iauno
       INNER JOIN acciones acuno ON iauno.id_accion = acuno.id
       WHERE acuno.id_proyecto = 3
       GROUP BY iauno.id_accion
       HAVING COUNT(*) > 1
     )), "-") as acc_convocadas,
  IFNULL((SELECT GROUP_CONCAT(id_accion)
      FROM inter_clientes_acciones iac
      WHERE iac.id_cliente = cl.id AND asistencia = 1 AND id_accion IN (
        SELECT iauno.id_accion
        FROM inter_clientes_acciones iauno
        INNER JOIN acciones acuno ON iauno.id_accion = acuno.id
        WHERE acuno.id_proyecto = 3
        GROUP BY iauno.id_accion
        HAVING COUNT(*) > 1)), "-") as acc_realizadas
FROM clientes cl
WHERE id_proyecto = 3

Answer 1

应该可以将table_cl.id的检查从子查询的最低级别移到更高级别，在那里（希望）MySQL会识别它。但是我有点担心子查询如何与id_accion分组的方式一起工作，而不会让它回归。使跟踪变得困难。

但是，最佳猜测是table_cl.id的检查位于子查询的最高级别。但是，如果没有表格声明，它是否真的可以进行任何测试。

SELECT table_cl.id, table_cl.nombre, table_cl.apellidos, (SELECT COUNT(*) FROM (
            SELECT iauno.id_accion, iauno.id_cliente, iauno.asistencia
            FROM inter_clientes_acciones iauno
            INNER JOIN clientes clientesuno ON clientesuno.id = iauno.id_cliente
            WHERE iauno.id_cliente IN(
                SELECT iasubuno.id_cliente
                FROM inter_clientes_acciones iasubuno
                INNER JOIN acciones acsubuno ON iasubuno.id_accion = acsubuno.id
                WHERE acsubuno.id_proyecto = 3
                GROUP BY iasubuno.id_accion 
                HAVING COUNT(*) = 1
            ) 
        )caa
    ) as convocado,
    (
        SELECT COUNT(*)  
        FROM inter_clientes_acciones iados
        INNER JOIN clientes cldos ON cldos.id = iados.id_cliente
        INNER JOIN 
        (
            SELECT iasubdos.id_cliente, 
                    id_accion
            FROM inter_clientes_acciones iasubdos
            INNER JOIN acciones acsubdos ON iasubdos.id_accion = acsubdos.id
            WHERE acsubdos.id_proyecto = 3 
            GROUP BY iasubdos.id_cliente, id_accion 
            HAVING COUNT(*) = 1
        ) sub0 ON iados.id_cliente = sub0.id_cliente
        WHERE iados.id_cliente = table_cl.id
    ) as realizado 
FROM clientes table_cl
WHERE table_cl.id_proyecto = 3 AND table_cl.id IN (
    SELECT iatres.id_cliente
    FROM inter_clientes_acciones iatres
    INNER JOIN clientes cltres ON cltres.id = iatres.id_cliente
    WHERE iatres.id_cliente IN(
        SELECT iasubtres.id_cliente
        FROM inter_clientes_acciones iasubtres
        INNER JOIN acciones acsubtres ON iasubtres.id_accion = acsubtres.id
        WHERE acsubtres.id_proyecto = 3
        GROUP BY iasubtres.id_accion 
        HAVING COUNT(*) = 1
        ) 
)

编辑

我已将其清理干净，但无法使用您的测试数据对其进行测试。我也把表名改为英语，这样我才能理解它们！

但是它确实执行了，而且我认为它符合你原本要做的事情。我觉得有可能进一步清理它，但我累了，感冒了，已经晚了！

SELECT table_cl.id, 
        table_cl.nombre, 
        table_cl.apellidos, 
        (
            SELECT COUNT(*)
            FROM client_actions 
            INNER JOIN
            (
                SELECT DISTINCT client_actions.id_cliente
                FROM client_actions 
                INNER JOIN actions ON client_actions.id_accion = actions.id
                WHERE actions.id_proyecto = 3
                GROUP BY client_actions.id_accion 
                HAVING COUNT(*) = 1
            ) sub1
            ON client_actions.id_cliente = sub1.id_cliente
        ) as convocado,
        (
            SELECT COUNT(*)
            FROM client_actions 
            INNER JOIN
            (
                SELECT DISTINCT client_actions.id_cliente
                FROM client_actions 
                INNER JOIN actions  ON client_actions.id_accion = actions.id
                WHERE actions.id_proyecto = 3 
                GROUP BY id_accion 
                HAVING COUNT(*) = 1
            ) sub2
            ON client_actions.id_cliente = sub2.id_cliente
            WHERE client_actions.id_cliente = table_cl.id
        ) as realizado 
FROM clients table_cl
INNER JOIN
(
    /* Get all the clients who have been the sole client for an action */
    SELECT DISTINCT client_actions.id_cliente
    FROM client_actions 
    INNER JOIN actions ON client_actions.id_accion = actions.id
    WHERE actions.id_proyecto = 3
    GROUP BY client_actions.id_accion 
    HAVING COUNT(*) = 1
) sub0
ON table_cl.id = sub0.id_cliente
WHERE table_cl.id_proyecto = 3

MySQL - 相关子查询失败，where子句中的列是unknwon

1 个答案: