Syntax
BIT_OR(expr)
Purpose
BIT_OR() returns the bitwise OR of all bits in expr.
The result is a binary string or a number. This depends on the type of the argument value. When the argument value contains a binary string and the argument is not a hexadecimal, bit, or NULL literal, a binary string result is calculated. Otherwise, a numeric result is calculated. If necessary, the function converts the argument value to an unsigned 64-bit integer.
If no matching row is found, BIT_OR() returns a neutral value whose all bits are set to 0. The neutral value has the same length as the argument value. NULL has no effect on the result unless all values are NULL. In this case, the result is a neutral value that has the same length as the argument value.
Examples
obclient> CREATE TABLE tbl1 (year YEAR (4), month INT(2) UNSIGNED ZEROFILL, day INT(2) UNSIGNED ZEROFILL);
Query OK, 0 rows affected
obclient> INSERT INTO tbl1 VALUES(2021,1,1),(2021,1,22),(2021,1,3),(2021,2,2), (2021,2,23),(2021,2,23);
Query OK, 6 rows affected
Records: 6 Duplicates: 0 Warnings: 0
obclient> SELECT * FROM tbl1;
+------+-------+------+
| year | month | day |
+------+-------+------+
| 2021 | 01 | 01 |
| 2021 | 01 | 22 |
| 2021 | 01 | 03 |
| 2021 | 02 | 02 |
| 2021 | 02 | 23 |
| 2021 | 02 | 23 |
+------+-------+------+
6 rows in set
obclient> SELECT year,month,BIT_COUNT(BIT_OR(1<<day)) AS days FROM tbl1 GROUP BY year,month;
+------+-------+------+
| year | month | days |
+------+-------+------+
| 2021 | 01 | 3 |
| 2021 | 02 | 2 |
+------+-------+------+
2 rows in set