Question

我最初在sql上有以下查询：

SELECT PROJECT_TYPE.PROJECT_TYPE_NAME, PROJECT.PROJECT_NAME, STATUS_TYPE.STATUS_TYPE_NAME,
COUNT(*) AS Tasks
FROM TASK
INNER JOIN PROJECT ON PROJECT.PROJECT_ID = TASK.PROJECT_ID
AND PROJECT.POURCENTAGE < 100               /* the following ANDs filters out inactive projects */
AND PROJECT.PROJECT_TYPE < 6
AND PROJECT.PROJECT_STATUS_ID < 3
INNER JOIN PROJECT_TYPE ON PROJECT.PROJECT_TYPE = PROJECT_TYPE.PROJECT_TYPE_ID
INNER JOIN STATUS_TYPE ON TASK.STATUS_TYPE_ID = STATUS_TYPE.STATUS_TYPE_ID
INNER JOIN PRIORITY_TYPE ON TASK.PRIORITY_TYPE_ID = PRIORITY_TYPE.PRIORITY_TYPE_ID
AND NOT PRIORITY_TYPE.PRIORITY_TYPE_NAME = 'Hold'                           
AND STATUS_TYPE.STATUS_TYPE_NAME IN ('To do','In Progress', 'To Verify')        /* only include projects in these 3 statuses */
GROUP BY PROJECT_TYPE.PROJECT_TYPE_NAME, PROJECT.PROJECT_NAME, STATUS_TYPE.STATUS_TYPE_NAME
order by 2, case status_type_name   /* orders by status */
  when 'To do' then 1
  when 'In Progress' then 2
  when 'To Verify' then 3
  else 4
  end

正确显示某个状态为such的任务计数。但是，我想替换

project.project_name

在另一个名为

的表中列出的较短描述的表

project.project_desc

，看起来像this。我试图通过用project.project_desc替换所有project.project_name并添加

来实现这一点。

AND PROJECT.PROJECT_DESC IS NOT NULL

在第一个内部联接下。但是，这会导致以下错误

SQL错误（306）：text，ntext和image数据类型不能比较或排序，除非使用IS NULL或LIKE运算符。

我现在不确定这有什么问题，所有的空值都应该被滤除，所以我很困惑为什么查询的运行方式不同于它之前做过，只是阅读一个不同的表。我无权编辑数据库，因此必须通过查询实现这些更改。

Answer 1

FWIW，我发现这更容易阅读...

SELECT pt.project_type_name
     , p.project_name
     , st.status_type_name
     , COUNT(*) tasks
  FROM task t
  JOIN project p  
    ON p.project_id = t.project_id
  JOIN project_type pt
    ON pt.project_type_id = p.project_type 
  JOIN status_type st
    ON st.status_type_id = t.status_type_id 
  JOIN priority_type xt
    ON xt.priority_type_id = task.priority_type_id 
 WHERE p.pourcentage < 100               
   AND p.project_type < 6
   AND p.project_status_id < 3
   AND xt.priority_type_name != 'hold'                           
   AND st.status_type_name IN('to do','in progress', 'to verify')  
 GROUP 
    BY pt.project_type_name
     , p.project_name
     , st.status_type_name
 ORDER 
    BY p.project_name
     , FIELD(status_type_name,'to do','in progress','to verify')

不确定是什么导致sql错误306

1 个答案: