[英]Returning Multiple Values of the same id in array
我正在從 MSSQL 2008 中的 3 個不同表中提取信息,我想獲得CC_qty
的 SUM 以及每個Location
壓縮為每個id
一個字段。 如果這可以在查詢本身中完成,那就太棒了 - listagg
和GROUP_CONCAT
不會削減它。 否則我一直在使用 array_reduce、array_merge、array_diff 都無濟於事。
這是我的查詢和原始數組:
SELECT a.id, a.qty, b.locationID, b.CC_qty, c.Location FROM (
SELECT left(id, 10) as id, MAX(qty) as qty
FROM db1
WHERE id like 'abc-abc%'
GROUP BY left(id, 10)
) as a
JOIN (
SELECT locationID, left(SKU, 10) as SKU, CC_qty FROM db2
WHERE CC_qty > 25
) as b on a.abc-abc = b.SKU
JOIN (
SELECT locationID, Location FROM db3
) as c on b.locationID = c.locationID
Array
(
[0] => Array
(
[id] => abc-abc-12
[qty] => 0
[locationID] => 276
[CC_qty] => 250
[Location] => NOP11
)
[1] => Array
(
[id] => abc-abc-12
[qty] => 0
[locationID] => 310
[CC_qty] => 1385
[Location] => NOP01
)
[2] => Array
(
[id] => abc-abc-23
[qty] => 0
[locationID] => 84
[CC_qty] => 116
[Location] => NOP06
)
[3] => Array
(
[id] => abc-abc-23
[qty] => 0
[locationID] => 254
[CC_qty] => 432
[Location] => NOP08
)
[4] => Array
(
[id] => abc-abc-23
[qty] => 0
[locationID] => 228
[CC_qty] => 101
[Location] => NOP04
)
[5] => Array
(
[id] => abc-abc-34
[qty] => 0
[locationID] => 254
[CC_qty] => 436
[Location] => NOP08
)
[6] => Array
(
[id] => abc-abc-34
[qty] => 0
[locationID] => 254
[CC_qty] => 62
[Location] => NOP08
)
[7] => Array
(
[id] => abc-abc-45
[qty] => 0
[locationID] => 75
[CC_qty] => 89
[Location] => NOP05
)
[8] => Array
(
[id] => abc-abc-45
[qty] => 0
[locationID] => 202
[CC_qty] => 372
[Location] => NOP07
)
)
這是我想要的輸出,為了簡單地知道我絕對需要哪些信息,我已經刪除了qty
和locationID
但那些不必刪除:
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
)
)
感謝您的關注!
由於我為 MySQL 留下了答案,因此它不會為此起作用。 我不太了解 MSSQL 無法使用它,所以這里有一種用 PHP 來完成它的方法,所以我不會讓你完全沒有答案。
$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
)
)
*/
我對 MSSQL 沒有任何經驗,但我非常有信心它提供了必要的合並、求和和連接功能。 無論如何,我不得不發布一個答案,因為我發現 Thomas 的答案不夠完善。
本質上,您應該使用id
值作為臨時鍵來確定您是在處理組的第一次出現還是后續出現。 在第一次遇到時,只需將整行保存到輸出數組。 對於屬於同一組的所有未來行,只需求和並連接所需的值。
要刪除結果數組中的臨時鍵,只需調用array_values($result)
。
代碼:(演示)
$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));
輸出:
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',
),
)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.