Mysql - Get Sum of All distinct rows based on other value - mysql

Question

I have a table which looks like this,
+-----+--------+-------+
| num | amount | value |
+-----+--------+-------+
| 1 | 12 | 1 |
| 1 | 12 | 1 |
| 2 | 13 | 1 |
| 4 | 15 | 0 |
| 2 | 13 | 1 |
| 3 | 14 | 1 |
| 3 | 14 | 1 |
| 1 | 12 | 1 |
+-----+--------+-------+
I want to sum the 'amount' column based on distinct 'num' column where 'value' is equal to 1,
for example, after running following query,
select DISTINCT num, amount, value from test where value =1 ;
distinct 'num' based table is
+-----+--------+-------+
| num | amount | value |
+-----+--------+-------+
| 1 | 12 | 1 |
| 2 | 13 | 1 |
| 3 | 14 | 1 |
+-----+--------+-------+
so i want the final result to be
12+13+14 = 39.
one more thing is that
i cant use subquery.
because it is already part of another query as well.
here is the script of my table is
SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for `test`
-- ----------------------------
DROP TABLE IF EXISTS `test`;
CREATE TABLE `test` (
`num` int(11) DEFAULT NULL,
`amount` int(11) DEFAULT NULL,
`value` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- ----------------------------
-- Records of test
-- ----------------------------
INSERT INTO `test` VALUES ('1', '12', '1');
INSERT INTO `test` VALUES ('1', '12', '1');
INSERT INTO `test` VALUES ('2', '13', '1');
INSERT INTO `test` VALUES ('4', '15', '0');
INSERT INTO `test` VALUES ('2', '13', '1');
INSERT INTO `test` VALUES ('3', '14', '1');
INSERT INTO `test` VALUES ('3', '14', '1');
INSERT INTO `test` VALUES ('1', '12', '1');

Answer 1

You can create a temp table containing distinct values and using SUM() to calculate amount. Your query will go like:
SELECT SUM(amount)
FROM (
SELECT DISTINCT t.`num`, t.`amount`
FROM test t
WHERE t.`value`=1
) temp

Answer 2

select num,sum(amount)/Count(*)
from test
where value = 1
group by num
order by num

Answer 3

I assume that every num has the same value (if not, how should you pick one?)
so, this will work:
SELECT SUM(DISTINCT amount)
FROM test
WHERE value = 1

Answer 4

SELECT SUM(amount) FROM (SELECT DISTINCT num, amount, VALUE FROM test WHERE VALUE =1) AS temp ;

Answer 5

Please check the solution
SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for `test`
-- ----------------------------
DROP TABLE IF EXISTS `test`;
CREATE TABLE `test` (
`num` int(11) DEFAULT NULL,
`amount` int(11) DEFAULT NULL,
`value` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- ----------------------------
-- Records of test
-- ----------------------------
INSERT INTO `test` VALUES ('1', '12', '1');
INSERT INTO `test` VALUES ('1', '12', '1');
INSERT INTO `test` VALUES ('2', '13', '1');
INSERT INTO `test` VALUES ('4', '15', '0');
INSERT INTO `test` VALUES ('2', '13', '1');
INSERT INTO `test` VALUES ('3', '14', '1');
INSERT INTO `test` VALUES ('3', '14', '1');
INSERT INTO `test` VALUES ('1', '12', '1');
-- ----------------------------
-- Insert the result set into temp table and then use sum().
-- ----------------------------
SELECT DISTINCT `num` , `amount`
INTO #tmp
FROM test
WHERE `value` =1
SELECT * FROM #tmp
SELECT sum(`amount`) as amount FROM #tmp