这是我从查询中获得的数组:
Array (
[0] => Array (
[0] => Array (
[PRODUCT] => ROSE
[VARIETY] => ADELE
[GOLD] => 160
[NORMAL] => 0
[TOTAL] => 160
)
[1] => Array (
[PRODUCT] => ROSE
[VARIETY] => ALESSO
[GOLD] => 1320
[NORMAL] => 550
[TOTAL] => 1870
)
[2] => Array (
[PRODUCT] => ROSE
[VARIETY] => ANASTACIA
[GOLD] => 440
[NORMAL] => 150
[TOTAL] => 590
)
[3] => Array (
[PRODUCT] => ROSE1
[VARIETY] => ANASTACIA1
[GOLD] => 420
[NORMAL] => 120
[TOTAL] => 540
)
[4] => Array (
[PRODUCT] => ROSE1
[VARIETY] => ANASTACIA1
[GOLD] => 440
[NORMAL] => 100
[TOTAL] => 540
)
[5] => Array (
[PRODUCT] => ROSE2
[VARIETY] => ANASTACIA2
[GOLD] => 640
[NORMAL] => 0
[TOTAL] => 640
)
[6] => Array (
[PRODUCT] => ROSE2
[VARIETY] => ANASTACIA2
[GOLD] => 440
[NORMAL] => 440
[TOTAL] => 880
)
)
)
)
键GOLD和NORMAL可以不同，具体取决于查询，但我想对GOLD、NORMAL 和 TOTAL by PRODUCT，类似这样，您可以省略 VARIETY:
Array(
[0] => Array(
[PRODUCT] => ROSE
[GOLD] => 1920
[NORMAL] => 700
[TOTAL] => 2620
)
[1] => Array(
[PRODUCT] => ROSE1
[GOLD] => 860
[NORMAL] => 220
[TOTAL] => 1080
)
[2] => Array(
[PRODUCT] => ROSE2
[GOLD] => 1080
[NORMAL] => 440
[TOTAL] => 1520
)
)
我尝试过这样的事情:
$harvest 是包含数据的数组
//This get the array_keys from the data
$arrayThead = array();
for ($i=0; $i < count($harvest) ; $i++) {
array_push($arrayThead, array_keys($harvest[$i][0]));
}
$arrayfoot= array();
foreach ($harvest as $key => $value) {
foreach ($value as $harv) {
foreach ($arrayThead as $key => $values) {
foreach ($values as $th) {
if($th != 'PRODUCT' && $th != 'VARIETY'){
$arrayfoot[$th] += $harv[$th];
}
}
}
}
}
但此时它会汇总所有产品的数据:
Array (
[GOLD] => 3850
[NORMAL] => 1360
[TOTAL] => 5220
)
已更新
这是 mysql 查询:
SELECT pr_products.product AS PRODUCT,
pr_varieties.variety AS VARIETY,
FORMAT(SUM(IF(pr_grades.grade='GOLD',pf_harvest.quantity,0)),0) AS GOLD,
SUM(IF(pr_grades.grade='NORMAL',pf_harvest.quantity,0)) AS NORMAL,
SUM(pf_harvest.quantity) AS TOTAL
FROM pf_harvest
INNER JOIN pf_performance ON pf_performance.id = pf_harvest.id_performance
INNER JOIN pr_products ON pr_products.id = pf_harvest.id_product
INNER JOIN pr_varieties ON pr_varieties.id = pf_harvest.id_variety
INNER JOIN pr_grades ON pr_grades.id = pf_harvest.id_grade
WHERE pf_performance.status = 100
AND pf_harvest.id_tenant = 1
AND pf_harvest.date = CURDATE()
GROUP BY pf_harvest.id_product, pf_harvest.id_variety
ORDER BY pf_harvest.id_product, pr_varieties.variety, pf_harvest.id_grade
如何像我解释的示例中那样将数据添加到数组中？ 谢谢!
最佳答案
<?php
$array = Array (
0 => Array (
0 => Array (
'PRODUCT' => 'ROSE',
'VARIETY' => 'ADELE',
'GOLD' => 160,
'NORMAL' => 0,
'TOTAL' => 160
),
1 => Array (
'PRODUCT' => 'ROSE',
'VARIETY' => 'ALESSO',
'GOLD' => 1320,
'NORMAL' => 550,
'TOTAL' => 1870
),
2 => Array (
'PRODUCT' => 'ROSE',
'VARIETY' => 'ANASTACIA',
'GOLD' => 440,
'NORMAL' => 150,
'TOTAL' => 590
),
3 => Array (
'PRODUCT' => 'ROSE1',
'VARIETY' => 'ANASTACIA1',
'GOLD' => 420,
'NORMAL' => 120,
'TOTAL' => 540
),
4 => Array (
'PRODUCT' => 'ROSE1',
'VARIETY' => 'ANASTACIA1',
'GOLD' => 440,
'NORMAL' => 100,
'TOTAL' => 540,
),
5 => Array (
'PRODUCT' => 'ROSE2',
'VARIETY' => 'ANASTACIA2',
'GOLD' => 640,
'NORMAL' => 0,
'TOTAL' => 640,
),
6 => Array (
'PRODUCT' => 'ROSE2',
'VARIETY' => 'ANASTACIA2',
'GOLD' => 440,
'NORMAL' => 440,
'TOTAL' => 880,
)
)
);
$totalByProduct = array();
foreach ($array as $items) {
foreach ($items as $item) {
if(!key_exists($item['PRODUCT'], $totalByProduct)){
$totalByProduct[$item['PRODUCT']] = $item;
continue;
}
$totalByProduct[$item['PRODUCT']]['GOLD'] += $item['GOLD'];
$totalByProduct[$item['PRODUCT']]['NORMAL'] += $item['NORMAL'];
$totalByProduct[$item['PRODUCT']]['TOTAL'] += $item['TOTAL'];
}
}
var_dump($totalByProduct);
?>
关于php - 通过参数 php 对数组中的数据求和，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43143337/