PHP按日期从数组中获取不同的ID数

Question

Looking for the simplest way to achieve the following in php:

I have this array with dates and ids

$data = array(
  array("date" => "2016-01", "ids" => array(1,2,3,4,5,6,7,8)),
  array("date" => "2016-02", "ids" => array(1,2,9,10,11,12)),
  array("date" => "2016-03", "ids" => array(3,16,17,18,19,20,21)),
  array("date" => "2016-04", "ids" => array(1,2,3,19,20,22,23))
);

The idea is to count ids by date but also return count(existing ids) since the start of every month separately. (I hope this is understandable)

The returned array should look like this:

$data = array(
  array("date" => "2016-01", "counter" => array(8)),
  array("date" => "2016-02", "counter" => array(2,4)), /* 2 existing ids from 2016-01 and 4 new in 2016-02) */
  array("date" => "2016-03", "counter" => array(1,0,6)), /* 1 existing from 2016-01 and 0 exiting from 2016-02 and 6 new */
  array("date" => "2016-04", "counter" => array(3,0,2,2)) /* etc. */
);



        | 2016-01 | 2016-02 | 2016-03 | 2016-04
------  | ------- | ------- | ------- | -------
2016-01 |     8   |         |         |         
------  | ------- | ------- | ------- | -------
2016-02 |     2   |    4    |         |        
------  | ------- | ------- | ------- | -------
2016-03 |     1   |    0    |    6    |        
------  | ------- | ------- | ------- | -------
2016-04 |     3   |    0    |    2    |    2   

Of course if there is a way to do that directly in sql i'll take it :)

Answer 1

Here is one way to do it. (I'm not sure if it's the simplest). Looping over the set of data is the easiest part.

Within that, consider each id from the current day with while ($id = array_pop($day['ids'])) . For each of those ids, start at the first day, and look for the id in that days set of ids. If it's found, increment the count for that day and continue with the next id ( continue 2 continues the while loop.) If it's not found, move on to the next day. If it's not found when you get to the current day (indicated by $i < $key in the for loop) then increment the count for the current day and move on to the next id.

foreach ($data as $key => $day) {
    $counts = array_fill(0, $key + 1, 0);
    while ($id = array_pop($day['ids'])) {
        for ($i=0; $i < $key; $i++) {
            $past_day = $data[$i];
            if (in_array($id, $past_day['ids'])) {
                $counts[$i]++;
                continue 2;
            }
        }
        $counts[$key]++;
    }
    $new_data[] = ['date' => $day['date'], 'counter' => $counts];
}

ps I have no idea how to do this in SQL.

Answer 2

You can do something like this to achieve the desired result,

$newArray = array();
$count = count($data);
for($i = 0; $i < $count; ++$i){
    $newArray[$i] = array('date' => $data[$i]['date'], 'counter' => array());
    for($j = 0; $j < $i; ++$j){
        $counter = 0;
        foreach($data[$i]['ids'] as $key => $id){
            if(in_array($id, $data[$j]['ids'])){
                ++$counter;
                unset($data[$i]['ids'][$key]);
            }
        }
        $newArray[$i]['counter'][] = $counter;
    }
    $newArray[$i]['counter'][] = count($data[$i]['ids']);
}

// display $newArray array
var_dump($newArray);

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Here's the live demo