MySQL SUM returns unexpected value - mysql

Question

I've been trying to use SUM() to total up some numeric values in my database, but it seems to be returning unexpected values. Here is the PHP-side for generating the SQL:
public function calcField(string $field, array $weeks)
{
$stmt = $this->conn->prepare('SELECT SUM(`'. $field .'`) AS r FROM `ws` WHERE `week` IN(?);');
$stmt->execute([implode(',', $weeks)]);
return $stmt->fetch(\PDO::FETCH_ASSOC)['r'];
}
Let's give it some example data for you folks at home:
$field = 'revenue';
$weeks = [14,15,16,17,18,19,20,21,22,23,24,25,26];
This returns this value:
4707.92
Without seeing the data, this may have seemed to have worked, but here's the rows for those weeks:
+----+------+------+---------+-------+---------+---------+------+----------+
| id | week | year | revenue | sales | gpm_ave | uploads | pool | sold_ave |
+----+------+------+---------+-------+---------+---------+------+----------+
| 2 | 14 | 2019 | 4707.92 | 292 | 13 | 0 | 1479 | 20 |
| 3 | 15 | 2019 | 4373.32 | 304 | 13 | 0 | 1578 | 19 |
| 4 | 16 | 2019 | 4513.10 | 275 | 14 | 0 | 1460 | 19 |
| 5 | 17 | 2019 | 4944.80 | 336 | 14 | 0 | 1642 | 20 |
| 6 | 18 | 2019 | 4343.87 | 339 | 13 | 0 | 1652 | 21 |
| 7 | 19 | 2019 | 3918.59 | 356 | 14 | 0 | 1419 | 25 |
| 8 | 20 | 2019 | 4091.20 | 247 | 19 | 0 | 1602 | 15 |
| 9 | 21 | 2019 | 4177.22 | 242 | 12 | 0 | 1588 | 15 |
| 10 | 22 | 2019 | 3447.88 | 227 | 18 | 0 | 1585 | 14 |
| 11 | 23 | 2019 | 3334.18 | 216 | 15 | 0 | 1675 | 13 |
| 12 | 24 | 2019 | 4736.15 | 281 | 13 | 0 | 1388 | 20 |
| 13 | 25 | 2019 | 4863.84 | 252 | 12 | 0 | 1465 | 17 |
| 14 | 26 | 2019 | 4465.95 | 281 | 21 | 0 | 1704 | 16 |
+----+------+------+---------+-------+---------+---------+------+----------+
As you can see, the total should be far greater than 4707.92 - and I notice that the first row revenue = 4707.92.
Here's what things get weird, if I add this into the function:
echo 'SELECT SUM(`'. $field .'`) AS r FROM `ws` WHERE `week` IN('. implode(',', $weeks) .');';
Which outputs:
SELECT SUM(revenue) AS r FROM ws WHERE week IN(14,15,16,17,18,19,20,21,22,23,24,25,26);
Copying and pasting this into MySQL CLI returns:
MariaDB [nmn]> SELECT SUM(revenue) AS r FROM ws WHERE week IN(14,15,16,17,18,19,20,21,22,23,24,25,26);
+----------+
| r |
+----------+
| 55918.02 |
+----------+
1 row in set (0.00 sec)
Which, looks a lot more accurate. However, that very same SQL statement returns the first row value rather than summing the column for those weeks.
This function gets triggered by an AJAX script:
$d = new Page\Snapshot\D();
# at the minute only outputting dump of values to see what happens
echo '<pre>'. print_r(
$d->getQuarterlySnapshot(new Page\Snapshot\S(), new App\Core\Date(), $_POST['quarter'], '2019'),
1
). '</pre>';
The function $d->getQuarterlySnapshot function:
public function getQuarterlySnapshot(S $s, Date $date, int $q, string $year)
{
switch($q)
{
case 1:
$start = $year. '-01-01 00:00:00';
$end = $year. '-03-31 23:59:59';
break;
case 2:
$start = $year. '-04-01 00:00:00';
$end = $year. '-06-30 23:59:59';
break;
case 3:
$start = $year. '-07-01 00:00:00';
$end = $year. '-09-30 23:59:59';
break;
case 4:
$start = $year. '-10-01 00:00:00';
$end = $year. '-12-31 23:59:59';
break;
}
$weeks = $date->getWeeksInRange('2019', 'W', $start, $end);
foreach ($weeks as $key => $week){$weeks[$key] = $week[0];}
return [
'rev' => $s->calcField('revenue', $weeks),
'sales' => $s->calcField('sales', $weeks),
'gpm_ave' => $s->calcField('gpm_ave', $weeks),
'ul' => $s->calcField('uploads', $weeks),
'pool' => $s->calcField('pool', $weeks),
'sold_ave' => $s->calcField('sold_ave', $weeks)
];
}
So I don't overwrite the value anywhere (that I can see at least). How do I use SUM() with the IN() conditional?

Answer 1

As pointed out in the comments by #MadhurBhaiya:
Comma separated week_id values is passing as a single string in the query: week_id in ('1,2,3...,15') . MySQL implicitly typecasts this to 1 and thus gets the first row only. You will need to change the query preparation code
Led me to breaking up the single ? into named parameters using a foreach loop:
public function calcField(string $field, array $weeks)
{
$data = [];
$endKey = end(array_keys($weeks));
$sql = 'SELECT SUM(`'. $field .'`) AS r FROM `ws` WHERE `week` IN(';
foreach ($weeks as $key => $week)
{
$sql .= ':field'. $key;
$sql .= ($key !== $endKey ? ',' : '');
$data[':field'. $key] = $week;
}
$sql .= ');';
$stmt = $this->conn->prepare($sql);
$stmt->execute($data);
return $stmt->fetch(\PDO::FETCH_ASSOC)['r'];
}
Now each numeric value is treated individually, which, is getting me the expected value.

Answer 2

Yes, it is showing only first row data in your query because you have mentioned where week IN (1,2,3,4,5,6,7,8,9,10,11,12,13); and your data shows that you have week 13 and 1 to 12 does not exist in your data from IN clause of your query.

Answer 3

You are selecting only the row with week 13, the week 1-12 is not existing in your data. That's why the result of your query is 43.900001525878906
Solution 1:
You might want to change the values in your IN()because you are filtering your data by the column week.
SELECT SUM(`revenue`) as `r` FROM `ws` WHERE `week` IN (13,14,15,16,17,18,19,20,21,22,23,24,25);
Solution 2:
You can change the week in your where clause to id
SELECT SUM(`revenue`) as `r` FROM `ws` WHERE `id` IN (1,2,3,4,5,6,7,8,9,10,11,12,13);