0

私はこのクエリを持っています:

SELECT
    p.name,
    COUNT(DISTINCT t.keyTask) AS totalTasksCount,
    SUM(DISTINCT CASE WHEN t.keyPriority = 24 AND (t.keyState = 16 OR t.keyState = 17) THEN 1 ELSE 0 END) AS highPriorityTasksCount,
    SUM(DISTINCT CASE WHEN t.keyState = 16 OR t.keyState = 17 THEN 1 ELSE 0 END) AS activesTasksCount,
    SUM(DISTINCT t.estimatedDuration) AS estimatedDuration,
    SUM(TIMESTAMPDIFF(SECOND,wp.start,wp.end)) * 1000 AS workedDuration
FROM projects_projects p
LEFT JOIN projects_tasks t
    ON p.keyProject = t.keyProject
LEFT JOIN projects_workPeriods wp
    ON wp.keyTask = t.keyTask
LEFT JOIN common_organizations o
    ON o.keyOrganization = p.keyOrganization
LEFT JOIN common_users uc
    ON uc.keyUser = p.keyUserCreator
LEFT JOIN common_users uu
    ON uu.keyUser = p.keyUserUpdater
GROUP BY
    p.keyProject
ORDER BY
    highPriorityTasksCount DESC,
    activesTasksCount DESC,
    p.updated DESC,
    p.name;

しかし、結果フィールドは、このクエリでは正常な 0 または 1 を返しますhighPriorityTasksCount。サブクエリなしでこれらのフィールドのケース結果値ではなく、行で作業activesTasksCountする方法があるかどうか疑問に思っていましたか?DISTINCTROW

現在の結果:

p.name,
totalTasksCount,
highPriorityTasksCount,
activesTasksCount,
estimatedDuration,
workedDuration

'Project 1',  '4', '1', '1',  '14400000',  '15300000'
'Project 2', '48', '1', '1',  '84600000', '503100000'
'Project 3',  '6', '1', '1', '108000000',        NULL
'Project 4',  '4', '1', '1',  '25200000',  '30600000'
'Project 5',  '5', '1', '1', '226800000',  '39600000'
'Project 6',  '2', '0', '1',        NULL,  '10800000'
'Project 7',  '9', '0', '1',        NULL,  '36900000'

期待される結果 ：

'Project 1',  '4', '1', '1',  '14400000',  '15300000'
'Project 2', '48','20', '2',  '84600000', '503100000'
'Project 3',  '6', '1', '1', '108000000',        NULL
'Project 4',  '4', '4', '2',  '25200000',  '30600000'
'Project 5',  '5', '5', '1', '226800000',  '39600000'
'Project 6',  '2', '0', '1',        NULL,  '10800000'
'Project 7',  '9', '0', '1',        NULL,  '36900000'

編集 ：

方法がないため、サブクエリを使用してクエリを変更しました。最適化に関するヘルプをいただければ幸いです。これは機能しています:

SELECT
    p.name,
    COUNT(DISTINCT t.keyTask) AS totalTasksCount,
    /* OLD SUM(DISTINCT CASE WHEN t.keyPriority = 24 AND (t.keyState = 16 OR t.keyState = 17) THEN 1 ELSE 0 END) AS highPriorityTasksCount, */
    (
        SELECT
    SUM(CASE WHEN st.keyPriority = 24 AND (st.keyState = 16 OR st.keyState = 17) THEN 1 ELSE 0 END)
    FROM projects_tasks st
    WHERE
    st.keyProject = p.keyProject
) AS highPriorityTasksCount,
    /* OLD SUM(DISTINCT CASE WHEN t.keyState = 16 OR t.keyState = 17 THEN 1 ELSE 0 END) AS activesTasksCount, */
    (
        SELECT
    SUM(CASE WHEN st.keyState = 16 OR st.keyState = 17 THEN 1 ELSE 0 END)
    FROM projects_tasks st
    WHERE
    st.keyProject = p.keyProject
) AS activesTasksCount,
    SUM(t.estimatedDuration) AS estimatedDuration,
    SUM(TIMESTAMPDIFF(SECOND,wp.start,wp.end)) * 1000 AS workedDuration
FROM projects_projects p
LEFT JOIN projects_tasks t
    ON p.keyProject = t.keyProject
LEFT JOIN projects_workPeriods wp
    ON wp.keyTask = t.keyTask
LEFT JOIN common_organizations o
    ON o.keyOrganization = p.keyOrganization
LEFT JOIN common_users uc
    ON uc.keyUser = p.keyUserCreator
LEFT JOIN common_users uu
    ON uu.keyUser = p.keyUserUpdater
GROUP BY
    p.keyProject
ORDER BY
    highPriorityTasksCount DESC,
    activesTasksCount DESC,
    p.updated DESC,
    p.name;
4

1 に答える 1

2

サブクエリなし。

いいえ、サブクエリを使用する必要があります。そうしないと、MySQLについて私が知らないことがあります。

于 2012-12-04T14:01:09.780 に答える