随工作日志条目返回重复行。我只需要最新的条目

Question

我做了这个有效的查询。问题是它显示每个工作日志的行。我只需要每个事件的最新工作日志。

SELECT DISTINCT
`HPD:Help Desk`.`Incident Number` AS Incident_Number,
`HPD:Help Desk`.`Status` AS Status,
`HPD:Help Desk`.`Priority` AS Priority,
`HPD:Help Desk`.`Assigned Group` AS Assigned_Group,
`HPD:Help Desk`.`Assignee` AS Assignee,
`HPD:Help Desk`.`Submit Date` AS Submit_Date,
`HPD:Help Desk`.`Last Modified Date` AS Last_Modified_Date,
`HPD:Help Desk`.`Description` AS Description,
`HPD:Help Desk`.`Submitter` AS Submitter,
`HPD:Search-Worklog`.`Worklog Description` AS Worklog_Description,
`HPD:Search-Worklog`.`Work Log ID` AS Work_Log_ID
FROM
`HPD:Help Desk` 
INNER JOIN `HPD:Search-Worklog` on (`HPD:Search-Worklog`.`InstanceId` = `HPD:Help Desk`.`InstanceId`)
WHERE 
(`HPD:Help Desk`.`Status` in ('Assigned','Pending','In Progress')
And `HPD:Help Desk`.`Submit Date` >= $__from/1000 and `HPD:Help Desk`.`Submit Date` <= $__to/1000)
ORDER BY
`HPD:Help Desk`.`Submit Date` ASC
LIMIT 20

我尝试过 HPD:Help Desk、HPD:Worklog 和 HPD:Search-Worklog 表单。我还尝试使用 CASE 语句仅显示每个事件的最新/最大工作日志 ID，但没有成功。我确信我错过了一些明显/简单的东西。

Answer 1

回答中

使用 row_number()