我已经看过了几个不同的答案,这个问题似乎并不能得到查询才能正常工作。Mysql的运行总计关闭一个表
这里是我的表有列用户,weekNo,salesTotalYTD。
我目前由周拉动这些出来,将它们分组,像这样:
+------+--------+---------------+
| user | weekNo | salesTotalYTD |
+------+--------+---------------+
|Jared | 1 | 200 |
+------+--------+---------------+
| Jim | 1 | 50 |
+------+--------+---------------+
|Jared | 2 | 30 |
+------+--------+---------------+
| Jim | 2 | 100 |
+------+--------+---------------+
我所试图做的,但不能完成如下:
+------+--------+---------------+
| user | weekNo | salesTotalYTD |
+------+--------+---------------+
|Jared | 1 | 200 |
+------+--------+---------------+
| Jim | 1 | 50 |
+------+--------+---------------+
|Jared | 2 | 230 |
+------+--------+---------------+
| Jim | 2 | 150 |
+------+--------+---------------+
这是查询我已经工作了第一遍但之后每次传球是错误的:
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
添但返回错误的更新代码礼貌:
$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);
}
你真的有'SELECT'出现两次这样呢? –
不,这是一个小姐副本,我纠正它。 –
我已提高您的问题。我们需要一个PHP人来看待这个问题并提出意见。我认为你使用会话变量的原始方法也很好。 –