Question

我在MSSQL 2008中从3个不同的表中提取信息，我希望每个CC_qty将Location和每个id的SUM压缩到一个字段中。如果这可以在查询本身中完成，那将是非常棒的 - listagg和GROUP_CONCAT并没有削减它。否则我一直在使用array_reduce，array_merge，array_diff无济于事。

这是我的查询和原始数组：

这是我想要的输出，为了简单了解我绝对需要哪些信息我已删除qty和locationID但不必删除这些信息：

 Array
    (
        [0] => Array
            (
                [id] => abc-abc-12
                [CC_qty] => 1635
                [Location] => NOP11, NOP01
            )

        [1] => Array
            (
                [id] => abc-abc-23
                [CC_qty] => 649
                [Location] => NOP06, NOP08, NOP04
            )

        [2] => Array
            (
                [id] => abc-abc-34
                [CC_qty] => 495
                [Location] => NOP08

        [3] => Array
            (
                [id] => abc-abc-45
                [CC_qty] => 461
                [Location] => NOP05, NOP07
            )

    )

感谢您的期待！

Answer 1

由于我为MySQL留下了答案，因此无法为此工作。我不太了解MSSQL以便使用它，所以这里有一种方法可以使用PHP，所以我不会完全没有答案。

$arr = array
(
    array
    (
        'id' => 'abc-abc-12',
        'qty' => 0,
        'locationID' => 276,
        'CC_qty' => 250,
        'Location' => 'NOP11'
    ),
    array
    (
        'id' => 'abc-abc-12',
        'qty' => 0,
        'locationID' => 310,
        'CC_qty' => 1385,
        'Location' => 'NOP01'
    ),
    array
    (
        'id' => 'abc-abc-23',
        'qty' => 0,
        'locationID' => 84,
        'CC_qty' => 116,
        'Location' =>  'NOP06'
    )
);

$combinedArr = array();

foreach ($arr as $a)
{
    $found = false;

    foreach ($combinedArr as $i => $b)
    {
        if ($b['id'] == $a['id'])
        {
            $found = true;
            $locs  = explode(',', $a['Location']);

            $combinedArr[$i]['CC_qty'] += $a['CC_qty'];

            if (!in_array($b['Location'], $locs))
            {
                $locs[] = $b['Location'];
                $combinedArr[$i]['Location'] = implode(', ', $locs);
            }
        }
    }

    if (!$found)
        $combinedArr[] = $a;
}

print_r($combinedArr);

/*
Array
(
    [0] => Array
        (
            [id] => abc-abc-12
            [qty] => 0
            [locationID] => 276
            [CC_qty] => 1635
            [Location] => NOP01, NOP11
        )

    [1] => Array
        (
            [id] => abc-abc-23
            [qty] => 0
            [locationID] => 84
            [CC_qty] => 116
            [Location] => NOP06
        )

)
*/

Answer 2

我没有使用MSSQL的经验，但是我很自信，它提供了合并，求和和连接所必需的功能。无论如何，我被迫发布答案，因为我发现Thomas的答案还不够完善。

本质上，应该使用id值作为临时键，以确定是要处理组的第一个匹配项还是随后的匹配项。第一次遇到时，只需将整行保存到输出数组即可。对于属于同一组的所有将来的行，只需求和并连接所需的值即可。

要删除结果数组中的临时键，只需调用array_values($result)。

代码：（Demo）

$array = [
    ['id' => 'abc-abc-12', 'qty' => 0, 'locationID' => 276, 'CC_qty' => 250, 'Location' => 'NOP11'],
    ['id' => 'abc-abc-12', 'qty' => 0, 'locationID' => 310, 'CC_qty' => 1385, 'Location' => 'NOP01'],
    ['id' => 'abc-abc-23', 'qty' => 0, 'locationID' => 84, 'CC_qty' => 116, 'Location' => 'NOP06'],
    ['id' => 'abc-abc-23', 'qty' => 0, 'locationID' => 254, 'CC_qty' => 432, 'Location' => 'NOP08'],
    ['id' => 'abc-abc-23', 'qty' => 0, 'locationID' => 228, 'CC_qty' => 101, 'Location' => 'NOP04'],
    ['id' => 'abc-abc-34', 'qty' => 0, 'locationID' => 254, 'CC_qty' => 436, 'Location' => 'NOP08'],
    ['id' => 'abc-abc-34', 'qty' => 0, 'locationID' => 254, 'CC_qty' => 62, 'Location' => 'NOP08'],
    ['id' => 'abc-abc-45', 'qty' => 0, 'locationID' => 75, 'CC_qty' => 89, 'Location' => 'NOP05'],
    ['id' => 'abc-abc-45', 'qty' => 0, 'locationID' => 202, 'CC_qty' => 372, 'Location' => 'NOP07'],
];

$result = [];
foreach ($array as $row) {
    if (!isset($result[$row['id']])) {
        $result[$row['id']] = $row;
    } else {
        $result[$row['id']]['qty'] += $row['qty'];                      // SUM
        $result[$row['id']]['locationID'] .= ", " . $row['locationID']; // CONCAT
        $result[$row['id']]['CC_qty'] += $row['CC_qty'];                // SUM
        $result[$row['id']]['Location'] .= ", " . $row['Location'];     // CONCAT
    }
}
var_export(array_values($result));

输出：

array (
  0 => 
  array (
    'id' => 'abc-abc-12',
    'qty' => 0,
    'locationID' => '276, 310',
    'CC_qty' => 1635,
    'Location' => 'NOP11, NOP01',
  ),
  1 => 
  array (
    'id' => 'abc-abc-23',
    'qty' => 0,
    'locationID' => '84, 254, 228',
    'CC_qty' => 649,
    'Location' => 'NOP06, NOP08, NOP04',
  ),
  2 => 
  array (
    'id' => 'abc-abc-34',
    'qty' => 0,
    'locationID' => '254, 254',
    'CC_qty' => 498,
    'Location' => 'NOP08, NOP08',
  ),
  3 => 
  array (
    'id' => 'abc-abc-45',
    'qty' => 0,
    'locationID' => '75, 202',
    'CC_qty' => 461,
    'Location' => 'NOP05, NOP07',
  ),
)

返回数组中相同id的多个值

2 个答案: