Group php array by date and sum another column's values - php

Question

I have an array:
Array
(
[0] => Array
(
[id] => 81
[placed] => 2013-09-19 16:32:53
[sub_total] => 786
)
[1] => Array
(
[id] => 80
[placed] => 2013-09-19 16:32:06
[sub_total] => 780
)
[2] => Array
(
[id] => 79
[placed] => 2013-09-18 17:06:48
[sub_total] => 786
)
[3] => Array
(
[id] => 78
[placed] => 2013-09-18 17:05:02
[sub_total] => 756
)
[4] => Array
(
[id] => 77
[placed] => 2013-09-17 17:02:53
[sub_total] => 786
)
[5] => Array
(
[id] => 76
[placed] => 2013-09-16 17:02:53
[sub_total] => 756
)
)
Is it possible to group this data by date and summarize subtotal amount to get output array:
Array
(
[0] => Array
(
[placed] => 2013-09-19
[sub_total] => 786 + 780
)
[2] => Array
(
[placed] => 2013-09-18
[sub_total] => 786 + 756
)
[3] => Array
(
[placed] => 2013-09-17 17:02:53
[sub_total] => 786
)
[4] => Array
(
[placed] => 2013-09-16 17:02:53
[sub_total] => 756
)
)

Answer 1

$output=array();
foreach($yourArray as $values)
{
$d=date("Y-m-d",strtotime($values["placed"]));
$output[$d]["sub_total"]+=$values["sub_total"];
}
print_r($output);
Fiddle
Credits: The initial array used on this fiddle was taken from the answer by Jason OOO below.

Answer 2

I tested this also: http://phpfiddle.org/main/code/rzv-ngp
<?php
Array
(
'0' => Array
(
'id' => '81',
'placed' => '2013-09-19 16:32:53',
'sub_total' => '786'
),
'1' => Array
(
'id' => '80',
'placed' => '2013-09-19 16:32:06',
'sub_total' => '780'
),
//...
);
$newarray = array();
foreach ($array as $value){
$temp = explode(" ", $value['placed']);
$date = $temp[0];
$total = (isset($newarray[$date]['sub_total']) ? $newarray[$date]['sub_total'] + $value['sub_total']: $value['sub_total']);
$newarray[$date] = array('placed' => $date, 'sub_total' => $total);
}
print_r($newarray);
?>

Answer 3

You can have that kind of array at querying time. Something like
select date_field_name,other_field from table_name group by Day(date_feild_name);
after that you can use a use foreach to work with each day's data!

Answer 4

Try using sub query
select field_name,DAY(date_field) as date,(select sum(sub_total) where DAY(date_field)=date) as sub_total from table_name group by day(date_field)

Answer 5

Try this code
<?php
$shop = array(array( id => '81',
placed => '2013-09-19',
sub_total => '786'
),
array( id =>'80',
placed => '2013-09-19',
sub_total => '780',
),
array( id => '79',
placed => '2013-09-18',
sub_total => '786'
),
array
(
id => '78',
placed => '2013-09-18',
sub_total => '756'
),
array(
id => '77',
placed => '2013-09-17',
sub_total => '786'
),
array(
id => '76',
placed => '2013-09-16',
sub_total => '756'
)
);
$result=array();
foreach($shop as $value)
{
if(!isset($result[$value['placed']]))
{
//echo $value['placed'];
//echo $result[$value['placed']];
$result[$value['placed']]=array('placed'=>$value['placed'],'sub_total'=>0);
}
$result[$value['placed']]['sub_total']+=$value['sub_total'];
}
print_r($result);
//print_r($shop);
?>