SQL - 另一个子查询

SELECT 
    * 
FROM 
    (
     SELECT 
      qu.job_id AS `qu_job_id` , qu.engineer_id, qu.id AS `quote_id` , jb.author, jb.job_id, jb.job_title, SUBSTRING(jb.job_description, 1, 200) AS `short_description` , jb.image_ref, jb.timestamp 
     FROM 
      jobs AS `jb` 
     LEFT JOIN 
      quotes AS `qu` 
     ON 
     qu.job_id = jb.job_id 
     WHERE 
     jb.author = " . ID . " 
     GROUP BY 
     jb.job_id 
     ORDER BY 
     jb.timestamp DESC 
    ) AS `jobs` 
    LEFT JOIN 
    (
     SELECT 
     COUNT(id) AS `total_replies` , job_id 
     FROM 
     quote_comments 
     WHERE 
     job_id = jobs.job_id 
    ) AS `replies` 
    ON 
     replies.job_id = jobs.job_id

我的问题使用结果从子查询是否有可能（以上不作为是工作）把这个结果从第一子查询，以防止在第二比较？（即job_id = jobs.job_id）SQL - 另一个子查询

问候

编辑

（这里作为评论有更多的空间）

SELECT 
    qu.engineer_id, 
    qu.id AS `quote_id`, 
    jb.author, 
    jb.job_id, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200) AS `short_description` , 
    jb.image_ref, 
    jb.timestamp, 
    count(qu.id) AS comment_count 
FROM 
    " . JOBS . " AS `jb` 
LEFT JOIN 
    " . QUOTES . " AS `qu` 
ON 
    qu.job_id = jb.job_id 
WHERE 
    jb.author = " . ID . " 
GROUP BY 
    jb.job_id, 
    qu.engineer_id, 
    qu.id, 
    jb.author, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200), 
    jb.image_ref, 
    jb.timestamp 
ORDER BY 
    jb.timestamp DESC

我最初的问题是不正确略。我实际上想要得到报价的数量，而不是报价评论。

上述建议在下面看起来似乎乍一看，但它返回的工作有多个报价submited×次数与每个报价计数为1。有任何想法吗？

来源

2012-01-27 Phil Jackson

是的，它是可能的。首先，你不能在子查询中使用ORDER BY。这是你得到的错误吗？ – 2012-01-27 20:06:42

它看起来像你试图得到一份工作，得到任何工作报价（或没有），以及作业的报价评论数。如果这是正确的，我想我知道答案。 – Hasteur 2012-01-27 20:11:37

我想要做的是从3个表中获取信息。所有这三个在comman，job_id和author_id中都有两个东西。我想要获得所有工作，并且还要获得1个查询中每个工作收到的评论数量。（p.s.顺序工作正常） – 2012-01-27 20:11:44

SELECT 
    qu.engineer_id, 
    qu.id AS `quote_id`, 
    jb.author, 
    jb.job_id, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200) AS `short_description` , 
    jb.image_ref, 
    jb.timestamp, 
    count(qc.id) AS comment_count 
FROM 
    jobs AS `jb` 
LEFT JOIN quotes AS `qu` ON qu.job_id = jb.job_id 
LEFT JOIN quote_comments AS `qc` ON qu.job_id = qc.job_id 
WHERE 
    jb.author = " . ID . " 
GROUP BY 
    jb.job_id, 
    qu.engineer_id, 
    qu.id, 
    jb.author, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200) 
    jb.image_ref, 
    jb.timestamp 
ORDER BY 
    jb.timestamp DESC

这将消除子查询解析并使您的总体查询运行得更快。我删除了报价工作ID，因为我们正在加入，因此只需要拉一次。我注意到我们直接从工作中拼接引用评论。您可能希望通过报价，让你看到有多少评论每帖有多少报价评论，而不是有整体的工作联系起来

来源

2012-01-27 20:23:21 Hasteur

感谢上述小周，并像一个魅力工作。谢谢 – 2012-01-27 20:32:09

等等，不，我不挑剔，因为它没有完全回答，我会在问题中解释更多。 – 2012-01-27 20:37:44

如果你将一个表'A'加入到2个表'B'和'C'中，这两个表都与A有一对多的关系，然后你用A.something'做一个'group，在集合函数中会得到错误的结果。 – 2012-01-27 23:16:25

SELECT 
    jb.author, 
    jb.job_id, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200) AS `short_description` , 
    jb.image_ref, 
    jb.timestamp, 
    count(qu.id) AS quote_count 
FROM 
    jobs AS `jb` 
LEFT JOIN quotes AS `qu` ON qu.job_id = jb.job_id 
WHERE 
    jb.author = " . ID . " 
GROUP BY 
    jb.job_id, 
    jb.author, 
    jb.job_title, 
    SUBSTRING(jb.job_description, 1, 200) 
    jb.image_ref, 
    jb.timestamp 
ORDER BY 
    jb.timestamp DESC

这将拉动由ID撰写，然后做的工作对提交的工作报价进行计数。 Unfortunateley获得报价作业的数量，你必须从报价表

来源

2012-01-27 20:48:00 Hasteur

编辑您的第一个查询离开过工程师ID和报价ID：

SELECT 
    * 
FROM 
    ( SELECT 
      qu.job_id AS `qu_job_id` , qu.engineer_id, qu.id AS `quote_id`, 
      jb.author, jb.job_id, jb.job_title, 
      SUBSTRING(jb.job_description, 1, 200) AS `short_description`, 
      jb.image_ref, jb.timestamp 
     FROM 
      jobs AS `jb` 
     LEFT JOIN 
      quotes AS `qu` 
     ON 
      qu.job_id = jb.job_id 
     WHERE 
      jb.author = " . ID . " 
     GROUP BY 
      jb.job_id 
           --- ORDER BY   --- not needed 
           --- jb.timestamp DESC --- here 
    ) AS `jobs` 
    LEFT JOIN 
    (
     SELECT 
      COUNT(*) AS `total_replies` , job_id 
     FROM 
      quote_comments 
           --- WHERE 
           --- job_id = jobs.job_id 
     GROUP BY    --- replacing the WHERE 
      job_id    --- with a GROUP BY 
    ) AS `replies` 
    ON 
     replies.job_id = jobs.job_id 
ORDER BY      --- order clause 
    timestamp DESC    --- moved here

来源

2012-01-27 21:21:28

SQL - 另一个子查询

回答

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