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MySQL how to fill missing dates in range?

Posted by: admin November 1, 2017 Leave a comment

Questions:

I have a table with 2 columns, date and score. It has at most 30 entries, for each of the last 30 days one.

date      score
-----------------
1.8.2010  19
2.8.2010  21
4.8.2010  14
7.8.2010  10
10.8.2010 14

My problem is that some dates are missing – I want to see:

date      score
-----------------
1.8.2010  19
2.8.2010  21
3.8.2010  0
4.8.2010  14
5.8.2010  0
6.8.2010  0
7.8.2010  10
...

What I need from the single query is to get: 19,21,9,14,0,0,10,0,0,14… That means that the missing dates are filled with 0.

I know how to get all the values and in server side language iterating through dates and missing the blanks. But is this possible to do in mysql, so that I sort the result by date and get the missing pieces.

EDIT: In this table there is another column named UserID, so I have 30.000 users and some of them have the score in this table. I delete the dates every day if date < 30 days ago because I need last 30 days score for each user. The reason is I am making a graph of the user activity over the last 30 days and to plot a chart I need the 30 values separated by comma. So I can say in query get me the USERID=10203 activity and the query would get me the 30 scores, one for each of the last 30 days. I hope I am more clear now.

Answers:

MySQL doesn’t have recursive functionality, so you’re left with using the NUMBERS table trick –

  1. Create a table that only holds incrementing numbers – easy to do using an auto_increment:

    DROP TABLE IF EXISTS `example`.`numbers`;
    CREATE TABLE  `example`.`numbers` (
      `id` int(10) unsigned NOT NULL auto_increment,
       PRIMARY KEY  (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=latin1;
    
  2. Populate the table using:

    INSERT INTO `example`.`numbers`
      ( `id` )
    VALUES
      ( NULL )
    

    …for as many values as you need.

  3. Use DATE_ADD to construct a list of dates, increasing the days based on the NUMBERS.id value. Replace “2010-06-06” and “2010-06-14” with your respective start and end dates (but use the same format, YYYY-MM-DD) –

    SELECT `x`.*
      FROM (SELECT DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY)
              FROM `numbers` `n`
             WHERE DATE_ADD('2010-06-06', INTERVAL `n`.`id` -1 DAY) <= '2010-06-14' ) x
    
  4. LEFT JOIN onto your table of data based on the time portion:

       SELECT `x`.`ts` AS `timestamp`,
              COALESCE(`y`.`score`, 0) AS `cnt`
         FROM (SELECT DATE_FORMAT(DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY), '%m/%d/%Y') AS `ts`
                 FROM `numbers` `n`
                WHERE DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY) <= '2010-06-14') x
    LEFT JOIN TABLE `y` ON STR_TO_DATE(`y`.`date`, '%d.%m.%Y') = `x`.`ts`
    

If you want to maintain the date format, use the DATE_FORMAT function:

DATE_FORMAT(`x`.`ts`, '%d.%m.%Y') AS `timestamp`

Questions:
Answers:

You can accomplish this by using a Calendar Table. That’s a table which you create once and fill with a date range (e.g. one dataset for each day 2000-2050; that depends on your data). Then you can make an outer join of your table against the calendar table. If a date is missing in your table, you return 0 for the score.