首页
学习
活动
专区
工具
TVP
发布
社区首页 >问答首页 >如何显示时间戳之间超过24小时的时间差

如何显示时间戳之间超过24小时的时间差
EN

Stack Overflow用户
提问于 2018-06-07 23:23:52
回答 1查看 195关注 0票数 5

我有一个将速度返回给响应的查询。除了timediff计算结果超过24小时的情况外,它可以正常工作。我得到一个错误"java.sql.SQLException:值中java.sal.Time类型的非法小时值'31‘“

'31‘的数量随时间戳之间小时数的变化而变化。

我不知道下一步该从哪里开始。我尝试过timestampdiff(),但在尝试运行查询时遇到错误。

如有任何意见或建议,我们将不胜感激。

代码语言:javascript
复制
SELECT a.*
FROM
(SELECT
databunker.purchases.id                                           AS 'pur_id',
 databunker.purchases.business                                    AS 'pur_business',
 databunker.purchases.time                                        AS 'pur_time',
 databunker.customers.id                                          AS 'customer_id',
 databunker.customers.phone_number                                AS 'customer_phone_#',
 databunker.customers.is_primary                                  AS 'customer_is_primary',
 map.object_salesforce_id,
 databunker.five9_calls.campaign_name,
 databunker.five9_calls.start_timestamp AS 'start_of_call',
 TIMEDIFF(databunker.five9_calls.start_timestamp, databunker.purchases.time) AS 'speed_to_response'

FROM databunker.purchases
 LEFT OUTER JOIN databunker.customers ON (databunker.purchases.customer_id = databunker.customers.id)
 LEFT OUTER JOIN (SELECT
                    databunker.mappings.object_salesforce_id,
                    databunker.mappings.object_id AS 'map_customer_id'
                  FROM databunker.mappings
                  WHERE databunker.mappings.object_class = 'customer') AS map
   ON (databunker.purchases.customer_id = map.map_customer_id)
 LEFT OUTER JOIN databunker.five9_calls ON (map.object_salesforce_id = 
databunker.five9_calls.salesforce_id)
WHERE databunker.purchases.business = 'uma'
     AND databunker.purchases.outcome_type = 'accepted'
     AND databunker.purchases.time >= curdate() - 1

     AND databunker.five9_calls.start_timestamp >= databunker.purchases.time

GROUP BY
 databunker.purchases.id,
 databunker.purchases.business,
 databunker.purchases.time,
 databunker.customers.id,
 databunker.customers.phone_number,
 map.object_salesforce_id,
 databunker.five9_calls.campaign_name,
 databunker.five9_calls.start_timestamp,
 TIMEDIFF(databunker.five9_calls.start_timestamp, databunker.purchases.time)

 ORDER BY databunker.purchases.id, databunker.purchases.time, 
 databunker.five9_calls.start_timestamp ASC) a

INNER JOIN

(
SELECT
  databunker.purchases.id AS 'pur_id_2',
  MIN(databunker.five9_calls.start_timestamp) AS 'call_time'


FROM databunker.purchases
  LEFT OUTER JOIN databunker.customers ON (databunker.purchases.customer_id = databunker.customers.id)
  LEFT OUTER JOIN (SELECT
                     databunker.mappings.object_salesforce_id,
                     databunker.mappings.object_id AS 'map_customer_id'
                   FROM databunker.mappings
                   WHERE databunker.mappings.object_class = 'customer') AS map
    ON (databunker.purchases.customer_id = map.map_customer_id)
  LEFT OUTER JOIN databunker.five9_calls ON (map.object_salesforce_id = databunker.five9_calls.salesforce_id)
WHERE databunker.purchases.business = 'uma'
      AND databunker.purchases.outcome_type = 'accepted'
      AND databunker.purchases.time >= curdate() - 1

      AND databunker.five9_calls.start_timestamp >= databunker.purchases.time

GROUP BY
  databunker.purchases.id
) b

ON a.pur_id = b.pur_id_2 AND a.start_of_call = b.call_time

EN

回答 1

Stack Overflow用户

发布于 2018-06-12 09:13:57

有“一天中的时间”和“时间跨度”。

MySQL的TIMEDIFF()返回的“时间跨度”为31小时,等等。

您的Java代码需要一个"time of day“,当然,这个值不应该超过23:59:59 (忽略闰秒)。

一种解决方法是

代码语言:javascript
复制
TO_SECONDS(databunker.five9_calls.start_timestamp) -
TO_SECONDS(databunker.purchases.time)

但它不会返回(例如) "31:58:47";相反,它将返回"115127“。

另一种解决方法是避免使用java.sal.Time,而是找到一个可以处理“时间跨度”的库。

票数 2
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/50744961

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档