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也许有点奇怪的标题描述，但我基本上想实现该GROUP_CONCAT()功能所做的事情，然后才保留双重条目。

我有四个要加入的表client，, doctor,physio和records

根据$client我想获取客户详细信息的变量，主治医生和治疗师（三个表中的一行）并加入该用户的所有记录。

说在这种情况下$client = 1。该records表有五个记录所在的列r_client_id = 1。如果我运行如下查询，我只会从records表中获取一条记录，即第一次出现 where r_client_id = 1（这当然是有道理的）：


SELECT
    client.c_id, client.c_name
    doctor.d_name, 
    physio.p_name, 
    records.r_record
FROM
    adm_clients AS client

    INNER JOIN 
        norm_client_doctor AS ncd ON ncd.ncd_client_id = client.c_id
    INNER JOIN 
        adm_doctor AS doctor ON doctor.d_id = ncd.ncd_doctor_id
    INNER JOIN 
        norm_client_physio AS ncp ON ncp.ncp_client_id = client.c_id
    INNER JOIN 
        adm_physio AS physio ON physio.p_id = ncp.ncp_physio_id

        LEFT JOIN 
            adm_doctor_records AS records ON records.r_client_id = client.c_id
WHERE
    client.c_id = '".$client."'

现在假设五个记录r_client_id = 1是这样的：


+------+-------------+-------------------+----------+
| r_id | r_client_id | r_record          | r_date   |
+------+-------------+-------------------+----------+
| 1    | 1           | regular visit     | 10/10/12 |
+------+-------------+-------------------+----------+
| 3    | 1           | emergency control | 24/10/12 |
+------+-------------+-------------------+----------+
| 7    | 1           | regular visit     | 08/09/12 |
+------+-------------+-------------------+----------+
| 18   | 1           | delivery          | 03/01/12 |
+------+-------------+-------------------+----------+
| 20   | 1           | health checkup    | 10/12/11 |
+------+-------------+-------------------+----------+

我希望我的输出是这样的数组：


Client 1
 - Name Doctor
 - Name Physio

Records
 - Emergency control, 24/10/12
 - Regular visit, 10/10/12
 - Regular visit, 08/09/12
 - Delivery, 03/01/12
 - Health checkup, 10/12/11

我能想到的最接近的一个是GROUP_CONCAT()在记录上添加一个，但是，当然，对“定期访问”进行分组，所以我会得到 4 行而不是 5


    GROUP_CONCAT(DISTINCT records.r_record SEPARATOR '|')
[..]
    echo(str_replace("|","<br>",$show->r_record));

有人知道如何显示所有匹配的记录吗？我觉得我很接近了，但我现在别无选择..

编辑： 我忘了提到当我删除时DISTINCT，它会显示所有记录两次..

解决了：

让它像这样工作：


    GROUP_CONCAT(DISTINCT 
        CONCAT (records.r_date, '~', records.r_record, '~', records.r_paraph)
    SEPARATOR '|') AS clientDoctorRecords,
4

1 回答 1

2

尝试：

SELECT
    client.c_id, client.c_name
    doctor.d_name, 
    physio.p_name, 
    GROUP_CONCAT(records.r_record)
FROM
    adm_clients AS client

    INNER JOIN 
        norm_client_doctor AS ncd ON ncd.ncd_client_id = client.c_id
    INNER JOIN 
        adm_doctor AS doctor ON doctor.d_id = ncd.ncd_doctor_id
    INNER JOIN 
        norm_client_physio AS ncp ON ncp.ncp_client_id = client.c_id
    INNER JOIN 
        adm_physio AS physio ON physio.p_id = ncp.ncp_physio_id

        LEFT JOIN 
            adm_doctor_records AS records ON records.r_client_id = client.c_id
WHERE
    client.c_id = '".$client."'
GROUP BY
    client.c_id

如果您想r_date在record一列中加入，那么您可以CONCAT先使用plain，然后GROUP_CONCAT对其进行操作。

于 2012-11-12T12:40:42.773 回答