如何在php中合并数组中的2个对象

Question

Array
(
    [month] => FEBRUARY
    [year] => 2018
    [org] => 40
    [action] => 4
)
Array
(
    [month] => FEBRUARY
    [year] => 2018
    [org] => 41
    [action] => 5
)

both have same content so how to merge this both so that i wil get data like this:

{"month":"FEBRUARY","year":"2018","org":"40,41","action":"4,5"}

Code:-

$query1 = $this->db->query($queryString);      
$children = array();
$yearArray = array(); 
foreach ($query1->result() as $data1)
{                               

    $yearArray['month'] = $data1->months;
    $yearArray['year'] = $data1->PAY_YEAR;                       
    $yearArray['org'] = $data1->org;                            
    $yearArray['action'] = $data1->action;                                
    print_r($result);
    array_push($children, $yearArray);
}

with above code i am getting this json but i want to change its format as i said earlier in question:

{"month":"FEBRUARY","year":"2018","org":"40","action":"4"},{"month":"MARCH","year":"2018","org":"40","action":"5"}

I want to change about output with this actually:

{"month":"FEBRUARY","year":"2018","org":"40,41","action":"4,5"}

Answer 1

Change your code like below:-

1.Only single array required

2.Use monthss_year as key (so that common values can merge easily)

3.Add common monthss_year org as comma seperated value

4.Add common monthss_year action as comma seperated value

5.From final array remove monthss_year keys and re-index it as numeric array (like 0,1,2,...)

6.Encode the array and print to see final result

Code need to be like below:-

$query1 = $this->db->query($queryString);      
$children = array(); // only single array required
foreach ($query1->result() as $data1)
{                               

    $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['month'] = $data1->months;
    $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['year'] = $data1->PAY_YEAR; 

    $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['org'] = (isset($children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['org'])) ? $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['org'].','.$data1->org : $data1->org;

    $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['action'] = (isset($children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['action'])) ? $children[trim($data1->monthss).'_'.trim($data1->PAY_YEAR)]['action'].','.$data1->action : $data1->action;
}

$children = array_values($children);

echo json_encode($children);

Note:- What data you shown for print_r($query1->result()) , for that my edited code is working perfectly fine:-

Output:- https://eval.in/960167

Answer 2

Try using

$query1 = $this->db->query($queryString);      
$children = array();

$temp = array();
foreach ($query1->result() as $data1)
{                          
    if(!isset($temp[$data1->PAY_YEAR.$data1->months]) {
       $temp[$data1->PAY_YEAR.$data1->months] = array();
    }  
    $yearArray = array();
    $yearArray['month'] = $data1->months;
    $yearArray['year'] = $data1->PAY_YEAR;                       
    $yearArray['org'] = $data1->org;                            
    $yearArray['action'] = $data1->action;                          

    $temp = array_merge_recursive($temp[$data1->PAY_YEAR.$data1->months], $yearArray);
}
print_r($temp);

Answer 3

You can try this, i think it will work.

$a = array("month"=>"FEBRUARY", "year"=>2018, "org"=>40, "action"=>4);
$b = array("month"=>"FEBRUARY", "year"=>2018, "org"=>41, "action"=>5);
$k = $m = 0;
foreach($a as $key => $val) {
  if($val == $b[$key] ) {
      if($b['org'] != $a['org'] && $k == 0) {
        $a['org'] = $a['org'].",".$b['org'];
        $k++;
      }
      if($b['action'] != $a['action'] && $m == 0) {
        $a['action'] = $a['action'].",".$b['action'];
        $m++;
      }
  }
}
print_r($a);

output:

Array ( [month] => FEBRUARY [year] => 2018 [org] => 40,41 [action] => 4,5 )