Compute dates and durations in mysql query

问题:

I have a sample table (table_name: track_task) like below:

task_id     stage        log_date
----------------------------------------
   3          1      2011-06-01 08:36:21
   9          1      2011-06-03 12:35:47
   3          2      2011-06-05 14:25:42
  21          1      2011-06-11 13:03:34
   9          2      2011-06-11 15:25:57
   3          3      2011-06-12 10:16:09
  21          2      2011-06-15 15:30:29
   3          4      2011-06-22 15:34:33
  21          3      2011-06-23 12:53:49
   9          4      2011-06-25 16:25:08

The data above is automatically populated when a task stage is progressed by some action in the application code. The stages run from 1 to 4. However, due to some logic, a task may skip stage 3. But all tasks end at stage 4. Probable task paths are like so:

(1,2,3,4) / (1,2,4) - Completed tasks
(1,2,3) / (1,2)     - In progress tasks

I need to query and retrieve a report that shows how long (in days) a task takes in each stage at a given time. I have come up with the following query so far:

SELECT z.task_id, a.log_date begin_date, d.log_date end_date, 
       DATEDIFF( b.log_date, a.log_date ) step1_days, 
       DATEDIFF( c.log_date, b.log_date ) step2_days, 
       DATEDIFF( d.log_date, c.log_date ) step3_days, 
       DATEDIFF( d.log_date, a.log_date ) cycle_days
FROM track_task z
LEFT JOIN track_task a ON ( z.task_id = a.task_id AND a.staging_id =1 )
LEFT JOIN track_task b ON ( z.task_id = b.task_id AND b.staging_id =2 )
LEFT JOIN track_task c ON ( z.task_id = c.task_id AND c.staging_id =3 )
LEFT JOIN track_task d ON ( z.task_id = d.task_id AND d.staging_id =4 )
GROUP BY z.oppty_id

to derive a result set like below:

task_id  begin_date   end_date   step1_days  step2_days  step3_days  cycle_days
-------------------------------------------------------------------------------
  3      2011-06-01  2011-06-22       4          7          10           21
  9      2011-06-03  2011-06-25       8         NULL       NULL          22
 21      2011-06-11     NULL          4          8         NULL         NULL

Is this a good way to go about it or there is a better way? How can I have the NULL values reported as zero? How can I retrieve then end_date for a task that is still in progress?

回答1:

That looks about How I would do something like this, assuming there's only going to ever be one instance of a stage used per task (if not, you need more work, both in defining your requirements, and writing the query).

To make null appear as 0, use the COALESCE() function:

SELECT z.task_id, COALESCE(DATEDIFF(b.log_date, a.log_date), 0) as step1_days

回答2:

No suggestions to improve the general approach to the problem. However, you can report NULL values as 0 (or anything else for that matter) using MySQL's COALESCE function http://dev.mysql.com/doc/refman/5.0/en/comparison-operators.html#function_coalesce.

Coalesce takes a list as input and returns the first non-null value.

For example:

     SELECT z.task_id, a.log_date begin_date, d.log_date end_date, 
           COALESCE(DATEDIFF( b.log_date, a.log_date ),0) step1_days, 
           COALESCE(DATEDIFF( c.log_date, b.log_date ),0) step2_days,
...

Will return 0 if step1_days or step2_days is NULL.

回答3:

About the third question, the following edit to the query resolves it.

SELECT z.task_id, a.log_date begin_date, e.log_date end_date, 
       COALESCE(DATEDIFF( b.log_date, a.log_date ),0) step1_days, 
       COALESCE(DATEDIFF( c.log_date, b.log_date ),0) step2_days, 
       COALESCE(DATEDIFF( d.log_date, c.log_date ),0) step3_days, 
       COALESCE(DATEDIFF( d.log_date, a.log_date ),0) cycle_days
FROM track_task z
LEFT JOIN track_task a ON ( z.task_id = a.task_id AND a.staging_id =1 )
LEFT JOIN track_task b ON ( z.task_id = b.task_id AND b.staging_id =2 )
LEFT JOIN track_task c ON ( z.task_id = c.task_id AND c.staging_id =3 )
LEFT JOIN track_task d ON ( z.task_id = d.task_id AND d.staging_id =4 )
LEFT JOIN track_task e ON ( z.task_id = e.task_id AND e.staging_id IN
    (
      SELECT max( staging_id )
      FROM z_table_trial
      GROUP BY oppty_id
    )
)
GROUP BY z.oppty_id