Need help constructing json with PHP

Question

I want the following json data created with php. I'am getting the required values from my database.

JSON i want to output(I created it manually to show):

{  
   "room":[  
      {  
         "id":"1",
         "temp":"20"
      },
      {  
         "id":"2",
         "temp":"30"
      }
   ]
}

Current PHP:

 $rows = array();
    if(!$allData->count()) {
        echo 'No results!';
    } else {
        foreach ($allData->results() as $allData) {
            $rows['id'] = $allData->id;
            $rows['data'] = $allData->temp;
        }
    }

// echo out as json data
    echo '{"rom":'.json_encode($rows).'}';

This is what my current PHP script outputs:

{"rom":{"id":"2","data":"20"}}

JSON formatted:

{  
   "rom":{  
      "id":"2",
      "temp":"30"
   }
}

As you can see it outputs only the last id and temp values..

I can't seem to find out what I'am doing wrong, any help is greatly appreciated.

MacGyer · Accepted Answer · 2015-11-22 21:02:46Z

3

You are missing one dimension in your array:

$rows = array();
if(!$allData->count()) {
    echo 'No results!';
} else {
    foreach ($allData->results() as $allData) {
        $rows[] = [
           'id' => $allData->id,
           'data' => $allData->temp,
        ];
    }
}

edited Nov 22, 2015 at 21:02

answered Nov 22, 2015 at 18:00

MacGyer

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1 Comment

5less Over a year ago

Also, please update do this: $rows[] = [ 'id' => $allData->id, 'data' => $allData->temp ]; You had too many double qoutes in there. Thank you for your help MacGyver!

Lipsyor · Accepted Answer · 2015-11-22 18:01:29Z

1

Consider this:

$rows[] = ["id"=>$allData->id, "data"=>$allData->temp];

I n this way you never overwrite previous data

answered Nov 22, 2015 at 18:01

Lipsyor

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