PHP Create Multidimension Array for JSON output

Question

I feel terrible even asking because I have been TRYING to understand and comprehend other peoples examples this is what i'm TRYING to accomplish

{
  "field": [
    {
      "appid": 0,
      "page": 0,
      "fieldname": "Sweet",
      "value": "Tammy Howell"
    },
    {
      "appid": 1,
      "page": 1,
      "fieldname": "Cecilia",
      "value": "Shana Jordan"
    },
    {
      "appid": 2,
      "page": 2,
      "fieldname": "Merritt",
      "value": "Copeland Pena"
    }
  ]
}

I need to be able to make the above JSON output happen when doing an SQL SELECT statement

Here is my currentCode

$x = 0;
$userFieldsResult = mysqli_query($database_id, "SELECT * FROM theDB.DynamicFields ORDER BY Page, Priority") or die (mysqli_error($database_id));

        if (mysqli_num_rows($userFieldsResult)<=0)
            {
                echo "nothing found";
                exit;
            } else 
            {
                while($row = mysqli_fetch_array($userFieldsResult))
                {
                    $userFields[$x]['appid'] = $row['appid'];
                    $userFields[$x]['page'] = $row['page'];
                    $userFields[$x]['fieldname'] = $row['fieldname'];
                    $userFields[$x]['value'] = $row['value'];

                    $x++;
                }
                echo json_encode($userFields);
                exit;
            }

echo json_encode($userFields);

This is normally how i just been outputting json so they each are just part of 1 array looping, but I am trying to understand how to create more "in-depth" JSON arrays so they have a specific identifier before they list out the information.

  [
    {
      "appid": 0,
      "page": 0,
      "fieldname": "Sweet",
      "value": "Tammy Howell"
    },
    {
      "appid": 1,
      "page": 1,
      "fieldname": "Cecilia",
      "value": "Shana Jordan"
    },
    {
      "appid": 2,
      "page": 2,
      "fieldname": "Merritt",
      "value": "Copeland Pena"
    }
  ]

In my example I just want to be able to have "field" be the "upper" part of the array that holds all the information as i showed in the example above.

Hassaan · Accepted Answer · 2016-01-26 08:53:54Z

4

You need to add field as parent array to your array.

Change from

echo json_encode($userFields);

Into

$json = array('fields' => $userFields);
echo json_encode($json);

edited Jan 26, 2016 at 8:53

answered Nov 4, 2015 at 4:46

Hassaan

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1 Comment

devpro Over a year ago

good catch, some more explanation would be beneficial for u.. :)

Ohgodwhy · Accepted Answer · 2015-11-04 04:46:16Z

1

In this particular configuration, you are being returned an object with properties, of which, the property's value may be an array.

So what you should actually be doing is creating a stdClass object in PHP and assigning it a property of "field" which is an empty array. Then you can push into this array stack.

$obj = new stdClass();
$obj->fields = array();
while(....){
    $obj->fields[$x]['appid'] = 
    $obj->fields[$x]['appid'] = $row['appid'];
    $obj->fields[$x]['page'] = $row['page'];
    $obj->fields[$x]['fieldname'] = $row['fieldname'];
    $obj->fields[$x]['value'] = $row['value'];
}

Now you can json encode the object and return it.

echo json_encode($obj);

answered Nov 4, 2015 at 4:46

Ohgodwhy

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