SQL - 排序左连接和主要结果答案

【问题标题】：SQL - ordering left joins and main resultsSQL - 排序左连接和主要结果
【发布时间】：2012-01-24 01:35:25
【问题描述】：

我目前正在尝试从数据库中返回所有工程师。除此之外，还有其他一些用户信息。但我希望 ac 表加入最近的行，然后按联系人姓名对主要结果进行排序。

谁能把我扔到正确的方向？

SELECT 
    mb.type, mb.contactname, mb.id, mb.inv_addressline2, mb.inv_addressline3, mb.signup_date,
    fb.engineer_id, fb.job_id, fb.neg_or_pos, fb.rating, fb.author_id,
    ac.engineer_id, ac.timestamp, ac.author_id,
    jb.job_id, SUBSTRING(jb.job_description, 0, 200) AS `short_description`, jb.timestamp 
FROM " . MEMBERS_TABLE . " AS mb 
LEFT JOIN " . ACCEPTED . " AS ac ON mb.id = ac.engineer_id 
LEFT JOIN " . FEEDBACK . " AS fb ON ac.job_id = fb.job_id 
LEFT JOIN " . JOBS . " AS jb ON fb.job_id = jb.job_id 
WHERE mb.type = 2 
GROUP BY mb.contactname 
ORDER BY ac.timestamp DESC

【问题讨论】：

标签： sql sql-order-by left-join

【解决方案1】：

最后想通了：

SELECT 
    res.type, res.contactname, res.id, res.inv_addressline2, res.inv_addressline3, res.signup_date,
    res.engineer_id_global, res.job_id_global, res.neg_or_pos, res.rating, res.author_id_global,
    res.timestamp_global, res.job_description               
FROM
    (
        SELECT
            mb.type, mb.contactname, mb.id, mb.inv_addressline2, mb.inv_addressline3, mb.signup_date,
            fb.engineer_id AS `engineer_id_global`, fb.timestamp AS `timestamp_global`, fb.job_id AS `job_id_global`, fb.neg_or_pos, fb.rating, fb.author_id AS `author_id_global`,
            ac.engineer_id, ac.timestamp, ac.author_id,
            jb.job_id, jb.job_description 
        FROM
            " . MEMBERS_TABLE . " AS mb 
        LEFT JOIN
            " . ACCEPTED . " AS ac 
        ON
            mb.id = ac.engineer_id 
        LEFT JOIN 
            " . FEEDBACK . " AS fb 
        ON 
            ac.job_id = fb.job_id 
        LEFT JOIN
            " . JOBS . " AS jb 
        ON 
             fb.job_id = jb.job_id 
        WHERE 
            mb.type = 2 
        ORDER BY
            fb.timestamp DESC
    ) AS res 
GROUP BY 
    res.contactname
ORDER BY
    res.contactname

干杯

编辑

以上方法现在有效！

【讨论】：

或许你也不应该提供建议？

【解决方案2】：

因此，仅从 ac 中选择时间戳为该组最大值的行；然后按联系人姓名排序。

这是一个查询，用于匹配每个工程师和她的最大时间戳：

SELECT engineer_id, max(timestamp) as latest
FROM ac
GROUP BY engineer_id;

然后只选择 (id,timestamp) 在此表中的行。

【讨论】：