Create new array from existing array with sub-array when value is repeating

Question

I have a single array. I'm posting as JSON because that's my API output.

[
  {
    "compId": 26,
    "addonId": "1",
    "addonRate": "32"
  },
  {
    "compId": 26,
    "addonId": "2",
    "addonRate": "51"
  },
  {
    "compId": 7,
    "addonId": "1",
    "addonRate": "11"
  },
  {
    "compId": 7,
    "addonId": "2",
    "addonRate": "12"
  }
]

And I want convert that array so that I can get the following output:

[
  {
    "addonId": "1",
    "companies": {
        "compId": 26,
        "addonRate": "32"
      },
      {
        "compId": 7,
        "addonRate": "11"
      }
  },
  {
    "addonId": "2",
    "companies": {
        "compId": 26,
        "addonRate": "51"
      },
      {
        "compId": 7,
        "addonRate": "12"
      }
  }
]

So far I'm able to extract distinct compId and addonId. But then I want add those companies in each addons and the respective rate in the sub array.

Any help would be great! Thanks

@SIDU I gave many try but so far as I said could only extract distinct compId and addonId into separate arrays. Any hint how do add the sub comp array into addon array with respective value? — Mehulkumar
– Mehulkumar, Commented Jan 20, 2016 at 5:20

wogsland · Accepted Answer · 2016-01-20 05:50:11Z

1

Have you tried something like

$newArray = array();
foreach ($startingArray as $row) {
    if (!isset($newArray[$row['addonId']])
        $newArray[$row['addonId']] = array(
            'addonId'=>$row['addonId'],
            'companies'=>array()
        );
    }
    $newArray[$row['addonId']]['companies'][] = array(   
        "compId"=> $row['compId'],
        "addonRate"=> $row['addonRate']
    );
}

edited Jan 20, 2016 at 5:50

answered Jan 20, 2016 at 5:20

wogsland

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2 Comments

Mehulkumar Over a year ago

HOLY! Just needed that simple classic if(!isset) statement. Just one correction "compId"=> $newArray[$row['compId']] would be "compId"=> $row['compId']. Thanks ALOT!

wogsland Over a year ago

Fixed. Glad I could help!

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