Question

我正试图从同一天收集一堆价值。那很好，但是我的问题是我想找到我在SUM中收集的值的ID。

我需要ID，因为我要提取一些数据到具有相同日期的ID。

这是我的桌子的一个例子：

╔════╦══════════╦═════════╦═════════╦═════════════╗
║ id ║ q_tap_1  ║ q_tap_2 ║ q_tap_3 ║    datex    ║
╠════╬══════════╬═════════╬═════════╬═════════════╣
║ 1  ║   -1.02  ║  0.00   ║  0.00   ║ 24.02.2019  ║
║ 2  ║   -23    ║  -1.23  ║  -354   ║ 24.02.2019  ║
║ 8  ║   -354   ║  -10.23 ║   -29   ║ 28.02.2019  ║ 
║ 4  ║   0.00   ║  -4.23  ║  -1.53  ║ 23.02.2019  ║
║ 7  ║   -34    ║  0.00   ║  0.00   ║ 23.02.2019  ║
║ 6  ║   -100   ║  -13.63 ║ -54.23  ║ 28.02.2019  ║
║ 9  ║   -1.32  ║  0.00   ║  0.00   ║ 28.02.2019  ║
║ 10 ║   -23    ║  0.00   ║  0.00   ║ 21.02.2019  ║
║ 11 ║   -5.23  ║  -40.53 ║ -8.32   ║ 21.02.2019  ║
║ 12 ║   -23    ║  -1.23  ║ -23.75  ║ 28.02.2019  ║ 
╚════╩══════════╩═════════╩═════════╩═════════════╝

这就是现在正在做的事情：

-349 (SUM of all the rows with the same date)
28.02.2019 (Date)
8 (Trying to get all the ids from the same day, but now its only showing the first)

-123
24.02.2019
1

-121
23.02.2019
4

-123
21.02.2019
10

这就是我要尝试的

-349
28.02.2019
8 6 9 12

-123
24.02.2019
1 2

-121
23.02.2019
4 7

-123
21.02.2019
10 11



$sql7 = 'SELECT DATE(datex) as datex, (SUM(q_tap_1) + SUM(q_tap_2) + SUM(q_tap_3) + SUM(q_tap_4) + SUM(q_tap_5) + SUM(q_tap_6)) AS tap, SUM(bonus_gevinst_1) AS gevinst, id FROM `test3` WHERE aktiv LIKE "Nei" GROUP BY DATE(datex) DESC';
$test7 = mysqli_query($conn,$sql7);

<?php
while ($test8 = mysqli_fetch_assoc($test7)) {

    $test32 = $test8['gevinst'] + $test8['tap'];
    echo '<form method="post">
            <input type="hidden" name="id" value="'.$test8['datex'].'">
            <li class="list-group-item text-secondary" style="border-color:#17a2b8;margin-bottom:5px;"><div style="display:-webkit-inline-box;"><p class="text-success" style="display: -webkit-inline-box;">'.$test32.'kr</p><br>
            <p class="text-info" style="display: -webkit-inline-box;">'.$test8['datex'].'</p>
            <p class="text-info" style="display: -webkit-inline-box;">'.$test8['id'].'</p>
            </li>
          </form>';
};
?>

Answer 1

您可以group_concat相关ID：

SELECT   DATE(datex) AS datex, 
         SUM(q_tap_1) + 
         SUM(q_tap_2) +
         SUM(q_tap_3) +
         SUM(q_tap_4) +
         SUM(q_tap_5) +
         SUM(q_tap_6) AS tap, 
         SUM(bonus_gevinst_1) AS gevinst,
         GROUP_CONCAT(id) AS ids -- Here
FROM     `test3` WHERE aktiv LIKE "Nei"
GROUP BY DATE(datex)
ORDER BY DATE(datex) DESC

SQL：获取属于SUM中不同值的正确ID

1 个答案: