Combine array of dates and sum values per month in PHP

Question

i've got an array of dates as keys and values (integers) in the form:

[2015-07-14] => 40
[2015-07-15] => 5
[2015-07-16] => 8
[2015-07-17] => 0
[2015-07-18] => 0
[2015-07-19] => 0
[2015-07-20] => 0
[2015-07-21] => 0
[2015-07-22] => 0
[2015-07-23] => 0
[2015-07-24] => 0
[2015-07-25] => 0
[2015-07-26] => 0
[2015-07-27] => 0
[2015-07-28] => 0
[2015-07-29] => 0
[2015-07-30] => 0
[2015-07-31] => 0
[2015-08-01] => 0
[2015-08-02] => 1
[2015-08-03] => 1
[2015-08-04] => 2
[2015-08-05] => 1

The startdate and enddate can be selected by the user.

Is there a quick and easy way to combine those dates and sum the values as per month? In my example, the result should look somethine like:

[2015-07] => 53
[2015-08] => 5

The way that I tried to solve that was to use explode functions and then try to recombine those, but that seems to me a bit more complicated than it needs to be.

What have you tried so far? We need to see what you've tried to tell you if it works; but the answer is likely to be subjective and as such this question is likely to be flagged for closure. — M1ke
– M1ke, Commented Sep 9, 2015 at 13:16
So quick question, if I was to select the starting point from 2015-07-22 to 2015-07-31 will I get a result of 0? Or will the days before 2015-07-22 count as days in the month, even though I haven't selected that day? Or better question can a user select days too or just months? — Andrei
– Andrei, Commented Sep 9, 2015 at 13:20
A combination of array_map and array_sum would be a solution — ahPo
– ahPo, Commented Sep 9, 2015 at 13:24
People answering with substr I just wanna point out it's a really bad idea as date formatting may change in the future. You'd want to stick with the DateTime for this one. — Andrei
– Andrei, Commented Sep 9, 2015 at 13:26

Pepo_rasta · Accepted Answer · 2015-09-09 13:24:09Z

2

something like

foreach($yourarray as $key=>$val){
   $result[substr($key,0,7)] += $val;
}

answered Sep 9, 2015 at 13:24

Pepo_rasta

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3 Comments

Yosha Over a year ago

Thank you very much. That works like a charm. I knew there was an easier way than using explode("-",...) :-)

Evert Over a year ago

Note that this will throw E_NOTICE errors if they are turned on (and they should).

Pepo_rasta Over a year ago

yes, it will throw notice, you can use @ to ignore it or just simply define each one, but PHP can easily handle theese ones, it wont harm your code read this question please: stackoverflow.com/questions/8132410/… its bad practice only because it would kill your code in another language but NOT in PHP

Alex M · Accepted Answer · 2015-09-09 13:33:29Z

0

Sure there is.

$per_month = array();
foreach ($arr as $key => $val) {
  $month = substr($key, 0, 7);
  if (!isset($per_month[$month])) {
    $per_month[$month] = 0;
  }
  $per_month[$month] += $val;
}
echo var_dump($per_month["2015-07"]);

Edit:

Also, instead of substr() you could use sscanf()

foreach...
list($y, $m, $d) = sscanf($key, "%d-%d-%d");
$per_month[$y.'-'.$m] += $val;
endforeach

... or explode()

foreach...
$var = explode('-', $key); // $var[0] = year, $var[1] = month ...
...
$per_month[$var[0].'-'.$var[1]] += $val;
endforeach

There are probably many more ways, these are just off the top of my head

edited Sep 9, 2015 at 13:33

answered Sep 9, 2015 at 13:23

Alex M

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1 Comment

Yosha Over a year ago

Thank you Alex, I was using the explode-function befor, but I thought this makes things more complicated than they need to be.

Khushal · Accepted Answer · 2021-12-16 18:24:00Z

0

For me substr() method didn't worked in Laravel.

So I tried it differently instead of substr() method.

$monthwise_data = array();
foreach($datewise_data as $key => $value){
    $month_key = date('Y-m', strtotime($key));
    if(array_key_exists($month_key, $monthwise_data)){
        // If we've already added this month to the new array, add the value
        $monthwise_data[$month_key] += $value;
    }
    else{
        // Otherwise create a new element with month as key and store the value
        $monthwise_data[$month_key] = $value;
    }
}
dd($monthwise_data);

answered Dec 16, 2021 at 18:24

Khushal

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