【发布时间】:2018-01-03 06:43:31
【问题描述】:
我查看了这个问题的几个不同答案,但似乎无法让查询正常工作。
这是我的表,其中包含列 user、weekNo、salesTotalYTD。
我目前正在将它们取出并按周分组,如下所示:
+------+--------+---------------+
| user | weekNo | salesTotalYTD |
+------+--------+---------------+
|Jared | 1 | 200 |
+------+--------+---------------+
| Jim | 1 | 50 |
+------+--------+---------------+
|Jared | 2 | 30 |
+------+--------+---------------+
| Jim | 2 | 100 |
+------+--------+---------------+
我正在尝试做但无法完成的事情如下:
+------+--------+---------------+
| user | weekNo | salesTotalYTD |
+------+--------+---------------+
|Jared | 1 | 200 |
+------+--------+---------------+
| Jim | 1 | 50 |
+------+--------+---------------+
|Jared | 2 | 230 |
+------+--------+---------------+
| Jim | 2 | 150 |
+------+--------+---------------+
这是我为第一次传递而工作的查询,但之后的每一次传递都是错误的:
SET @runtot:=0
SELECT
salesTotalYTD,
user,
(@runtot := @runtot + salesTotalYTD) AS rt
FROM weeksAndSalesmantbl
GROUP BY user, weekNo
ORDER BY (CASE WHEN weekNo = 52 THEN 0 ELSE 1 END) ASC, weekNo, user ASC
更新
由 Tim 提供的更新代码但返回错误:
$assignments = "
SELECT
t1.user,
t1.weekNo,
(SELECT SUM(t2.salesTotalYTD) FROM weeksAndSalesmantbl t2
WHERE t2.user = t1.user AND t2.weekNo <= t1.weekNo) AS salesTotalYTD
FROM weeksAndSalesmantbl t1
ORDER BY
t1.weekNo,
t1.user";
$salesTotalSalesManCumulative = [];
$assignmentsqry = mysqli_query($db,$assignments);
if (!$assignmentsqry) {
printf("Error: %s\n", mysqli_error($db));
exit();
}
while ($row = mysqli_fetch_array($assignmentsqry)) {
$float = floatval($row['salesTotalYTD']);
$float = round($float,2);
array_push($salesTotalSalesManCumulative,$float);
}
【问题讨论】:
标签: php mysql cumulative-sum