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  • 按缺陷类型分组查询激活数量、1级、2级BUG数量

    SELECT
    type,
    count( * ) AS 激活数量,
    sum( IF ( severity = 1, 1, 0 ) ) AS 1级数量,
    sum( IF ( severity = 2, 1, 0 ) ) AS 2级数量,
    sum( IF ( severity = 3, 1, 0 ) ) AS 3级数量,
    sum( IF ( severity = 4, 1, 0 ) ) AS 4级数量
    FROM
    zt_bug
    WHERE
    product = '189'
    AND STATUS = 'active'
    GROUP BY
    type

    #激活状态1级、2级BUG数量
    SELECT
    type,
    count( * ) AS 激活数量,
    sum( IF ( severity = 1, 1, 0 ) ) AS 1级数量,
    sum( IF ( severity = 2, 1, 0 ) ) AS 2级数量,
    sum( IF ( severity = 3, 1, 0 ) ) AS 3级数量,
    sum( IF ( severity = 4, 1, 0 ) ) AS 4级数量
    FROM
    zt_bug
    WHERE
    product = '189'
    AND STATUS = 'active'
    GROUP BY
    type


    #当日新创建、新解决和新关闭的数量
    SELECT
    sum( IF ( date_format( openedDate, '%Y-%m-%d' ) = date_format( NOW( ), '%Y-%m-%d' ), 1, 0 ) ) AS 今天创建,
    sum( IF ( date_format( resolvedDate, '%Y-%m-%d' ) = date_format( NOW( ), '%Y-%m-%d' ), 1, 0 ) ) AS 今天解决,
    sum( IF ( date_format( closedDate, '%Y-%m-%d' ) = date_format( NOW( ), '%Y-%m-%d' ), 1, 0 ) ) AS 今天关闭
    FROM
    zt_bug
    WHERE
    product = '189'

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  • 原文地址:https://www.cnblogs.com/ruijie/p/14958422.html
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