mysql - calculate difference between two dates - mysql

Question

I have a table which looks like below.
CREATE TABLE `table_growth` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`timestamp` datetime DEFAULT CURRENT_TIMESTAMP,
`table_name` varchar(50) DEFAULT NULL,
`rows` int(11) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=184 DEFAULT CHARSET=utf8
Example of rows in the table:
+-----+---------------------+--------------------------+-------+
| id | timestamp | table_name | rows |
+-----+---------------------+--------------------------+-------+
| 110 | 2019-03-01 06:00:00 | attachments | 640 |
| 111 | 2019-03-01 06:00:00 | contacts | 0 |
| 112 | 2019-03-01 06:00:00 | copy_menuitem_options | 3038 |
| 113 | 2019-03-01 06:00:00 | copy_menuitem_suboptions | 9779 |
| 114 | 2019-03-01 06:00:00 | copy_menuitems | 12118 |
| 115 | 2019-03-02 06:00:00 | attachments | 638 |
| 116 | 2019-03-02 06:00:00 | contacts | 0 |
| 117 | 2019-03-02 06:00:00 | copy_menuitem_options | 3039 |
| 118 | 2019-03-02 06:00:00 | copy_menuitem_suboptions | 9789 |
| 119 | 2019-03-02 06:00:00 | copy_menuitems | 12128 |
+-----+---------------------+--------------------------+-------+
I want to calculate the diff(rows) between 2 days. like date(timestamp)='2019-03-02' - date(timestamp)='2019-03-01'
Expected results
table_name | Rows Diff
------------------------------
attachments | 2
contacts | 0
copy_menuitem_options | 1
copy_menuitem_suboptions| 10
copy_menuitems | 10
I tried these queries, but somewhere its failing.
SELECT x.table_name
, (y.rows-x.rows)as diff
FROM dbadmin.table_growth x
JOIN dbadmin.table_growth y
ON y.id = x.id
AND DATE(y.timestamp) = '2019-03-02'
WHERE DATE(x.timestamp) = '2019-03-01';
select x.table_name, (y.rows - x.rows) as doff
from table_growth x join
table_growth y on y.id=x.id and DATE(y.timestamp) = '2019-03-02'
WHERE DATE(x.timestamp) = '2019-03-01';

Answer 1

Your second query is on the right track, but the join condition is partially off. You should be asserting that the table names, not ids, match:
SELECT
x.table_name,
(x.rows - y.rows) AS diff
FROM table_growth x
INNER JOIN table_growth y
ON x.table_name = y.table_name and
DATE(y.timestamp) = '2019-03-02'
WHERE
DATE(x.timestamp) = '2019-03-01';
Note: Your current output is slightly ambiguous, because it is not clear which rows value comes first in the difference, or if perhaps you want to report an absolute value.

Answer 2

this will work;
select distinct a.table_name,(a.rows-b.rows) diff from table_growth a,table_growth b
where a.table_name=b.table_name;

Answer 3

If you have only one row per date, then this might be the fastest approach:
SELECT g.table_name,
SUM(CASE WHEN DATE(g.timestamp) = '2019-03-02'
THEN g.rows
WHEN DATE(g.timestamp) = '2019-03-01'
THEN -g.rows
ELSE 0
END) as diff
FROM dbadmin.table_growth g
WHERE g.timestamp >= '2019-03-01' AND
g.timestamp < '2019-03-03'
GROUP BY g.table_name;
In particular, this can make use of an index on table_growth(timestamp, table_name, rows).