改进SQL查询

时间：2016-06-28 07:55:05

标签： mysql sql performance

我检索了一个由于SQL查询而需要很长时间才能运行的项目。

我想知道你是否知道如何改进它。 至少如果有办法改进它。

据我所知，查询返回1500行，并从其他表中生成SUM和COUNT。

SELECT
p.id,
p.datecreate,
p.title,
p.address,
p.address2,
p.code_postal,
p.ville,
p.description,
p.description_admin,
p.idstatut,
p.idstatut_admin,
p.reference,
p.star,
p.logo,
p.deleted,
cc.value as starcolor,
p.idsociete,
soc.nom as societe_nom,
s.titlestatut,
s.fontcolor,
s.label as label,
sa.titlestatut as titlestatut_admin,
sa.fontcolor as fontcolor_admin,
sa.label as label_admin,
(SELECT SUM(nbr) FROM plans as cplans WHERE cplans.idprojets = p.id) as count_commandes,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '1' AND cfiles.bat_valid = '0') as count_averifier,
(SELECT count(rowid) FROM files_modif_title as cfmt WHERE cfmt.idprojets = p.id AND cfmt.statut = '7' AND cfmt.hide = '0') as count_modif_nohide,
(SELECT count(rowid) FROM files_modif_title as cfmt WHERE cfmt.idprojets = p.id AND cfmt.statut = '7' AND cfmt.hide IN (1)) as count_modif_hide,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '2' AND cfiles.bat_valid = '0') as count_bat_attente,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '2' AND cfiles.bat_valid = '1') as count_bat_valider
FROM projets as p INNER JOIN societe AS soc ON p.idsociete = soc.id
INNER JOIN statuts AS s ON p.idstatut = s.id
INNER JOIN statuts AS sa ON p.idstatut_admin = sa.id
LEFT JOIN const AS cc ON cc.name = p.star AND cc.parent = 'star'
WHERE  p.idstatut IN (3)
AND p.deleted = 0
GROUP BY p.id
ORDER BY p.datecreate DESC

谢谢你们！

编辑-----

我做了什么，可能会更好吗？

SELECT
    p.id,
    p.datecreate,
    p.title,
    p.address,
    p.address2,
    p.code_postal,
    p.ville,
    p.description,
    p.description_admin,
    p.idstatut,
    p.idstatut_admin,
    p.reference,
    p.star,
    p.logo,
    p.deleted,
    cc.value as starcolor,
    p.idsociete,
    soc.nom as societe_nom,
    s.titlestatut,
    s.fontcolor,
    s.label as label,
    sa.titlestatut as titlestatut_admin,
    sa.fontcolor as fontcolor_admin,
    sa.label as label_admin,
    CC.count_commandes,
    CA.count_averifier,
    CMN.count_modif_nohide,
    CMH.count_modif_hide,
    CBA.count_bat_attente,
    CBV.count_bat_valider
FROM
    projets as p
        INNER JOIN societe AS soc
            ON p.idsociete = soc.id
        INNER JOIN statuts AS s
            ON p.idstatut = s.id
        INNER JOIN statuts AS sa
            ON p.idstatut_admin = sa.id
        LEFT JOIN const AS cc
            ON cc.name = p.star
            AND cc.parent = 'star'
        LEFT JOIN (

                SELECT idprojets, SUM(nbr) as count_commandes
                FROM plans
                GROUP BY idprojets
            ) AS CC
            ON p.id = CC.idprojets
        LEFT JOIN (

                SELECT idprojets, COUNT(*) AS count_averifier
                FROM files
                GROUP BY idprojets
                WHERE cfiles.folder = 1 AND cfiles.bat_valid = 0
            ) AS CA
            ON p.id = CA.idprojets
        LEFT JOIN (

                SELECT idprojets, COUNT(*) as count_modif_nohide
                FROM files_modif_title
                WHERE statut = 7 AND hide = 0
                GROUP BY idprojets
            ) AS CMN
            ON p.id = CMN.idprojets
        LEFT JOIN (

                SELECT idprojets, COUNT(*) as count_modif_hide
                FROM files_modif_title
                WHERE statut = 7 AND hide = 1
                GROUP BY idprojets
            ) AS CMH
            ON p.id = CMH.idprojets
        LEFT JOIN (

                SELECT idprojets, COUNT(*)
                FROM files
                WHERE folder = 2 AND bat_valid = 0
                GROUP BY idprojets
            ) AS CBA
            ON p.id = CBA.idprojets
        LEFT JOIN (

                SELECT idprojets, COUNT(*)
                FROM files
                WHERE folder = 2 AND bat_valid = 1
                GROUP BY idprojets
            ) AS CBV
            ON p.id = CBV.idprojets
WHERE
    p.idstatut IN (3)
    AND p.deleted = 0
GROUP BY p.id
ORDER BY p.datecreate DESC;

感谢您所说的，主要问题是SELECT子句中嵌入了6个SELECT。这些是针对应用程序放在一起的每条记录进行评估，因此它执行1500 x 6 = 9000个查询！ 通过这样做，我有9001个查询，现在只有7个，因为子查询在运行时只被评估一次。这是对的吗？

4 个答案:

答案 0 :(得分：2)

这是你想要改进的部分：

(SELECT SUM(nbr) FROM plans as cplans WHERE cplans.idprojets = p.id) as count_commandes,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '1' AND cfiles.bat_valid = '0') as count_averifier,
(SELECT count(rowid) FROM files_modif_title as cfmt WHERE cfmt.idprojets = p.id AND cfmt.statut = '7' AND cfmt.hide = '0') as count_modif_nohide,
(SELECT count(rowid) FROM files_modif_title as cfmt WHERE cfmt.idprojets = p.id AND cfmt.statut = '7' AND cfmt.hide IN (1)) as count_modif_hide,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '2' AND cfiles.bat_valid = '0') as count_bat_attente,
(SELECT count(id) FROM files as cfiles WHERE cfiles.idprojets = p.id AND cfiles.folder = '2' AND cfiles.bat_valid = '1') as count_bat_valider

您可以对此使用条件聚合，并仅将表连接一次：

count(nbr) as..,
count(CASE WHEN cfiles.folder = '1' and cfiles.bat_valid = '0' then id END) as ..,
count(CASE WHEN cfiles.folder = '2' and cfiles.bat_valid = '0' then id END) as ..,
count(CASE WHEN cfiles.folder = '2' and cfiles.bat_valid = '1' then id END) as ..,
........

添加联接

JOIN files cfiles
 ON(cfiles.idprojets = p.id)

对files_modif_title

执行完全相同的操作

答案 1 :(得分：1)

  1. 希望你的主表有索引
  2. 使用join
    3. 而不是使用子查询

答案 2 :(得分：1)

试，

SELECT
p.id,
p.datecreate,
p.title,
p.address,
p.address2,
p.code_postal,
p.ville,
p.description,
p.description_admin,
p.idstatut,
p.idstatut_admin,
p.reference,
p.star,
p.logo,
p.deleted,
cc.value as starcolor,
p.idsociete,
soc.nom as societe_nom,
s.titlestatut,
s.fontcolor,
s.label as label,
sa.titlestatut as titlestatut_admin,
sa.fontcolor as fontcolor_admin,
sa.label as label_admin,
count (case when cfiles.folder = '2' AND cfiles.bat_valid = '0' then 1 end ) count_bat_attente,
count (case when cfiles.folder = '1' AND cfiles.bat_valid = '0' then 1 end ) count_averifier,
count (case when cfiles.folder = '2' AND cfiles.bat_valid = '1' then 1 end) count_bat_valider,
SUM(nbr) as count_commandes,
count (case when cfmt.statut = '7' AND cfmt.hide = '0' then 1 end) count_modif_nohide,
count (case when cfmt.statut = '7' AND cfmt.hide IN (1) then 1 end) count_modif_hide

FROM projets as p INNER JOIN societe AS soc ON p.idsociete = soc.id
INNER JOIN statuts AS s ON p.idstatut = s.id
INNER JOIN statuts AS sa ON p.idstatut_admin = sa.id
left join files on files.idprojets=p.id
left join plans cplans on cplans.idprojets = p.id
left join files_modif_title cfmt on cfmt.idprojets = p.id
LEFT JOIN const AS cc ON cc.name = p.star AND cc.parent = 'star'
WHERE  p.idstatut IN (3)
AND p.deleted = 0
GROUP BY p.id
ORDER BY p.datecreate DESC

答案 3 :(得分：1)

回答你的问题“我在这里做了什么，可能会更好吗？”，......

尽管您的9001逻辑有效，但还有另一个问题...... LEFT JOIN ( SELECT ... )正在构建一些行数（1500？）的临时表。如果您在5.6之前运行版本，则这些tmp表没有索引，必须重复扫描。现在我们谈论1500 * 1500 * 6 =数百万次操作，而不仅仅是9001。

即使使用5.6，也有额外的步骤（6次）来发现最佳索引并构建它。

但是，你真正的问题是“我怎样才能加快速度？”。其他人已经很好地回答了这个问题。

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