How to create an array of objects In PHP using a while loop

Question

Hi so I am attempting to return all the "people" from my database and then create an object of the "person" data, then at the end of each while loop add the person to the array. Outside of the while loop i am then encoding said array.

Below attached is my current code:

class person {
   public $firstName = "";
   public $lastName = "";
}

$people = array();
$result = mysqli_query($con,"SELECT * FROM People");

while($row = mysqli_fetch_array($result)) {
      $firstName=$row['firstName'];
      $lastName=$row['lastName'];

      $e = new Person();        
      $e->firstName  = $firstName;
      $e->lastName = $lastName;

      array_push($people, $e);
}

json_encode($people);

Many thanks

And are you having any particular problem?

Mark Baker
– Mark Baker

2014-10-27 19:03:11 +00:00
Commented Oct 27, 2014 at 19:03 — Mark Baker
– Mark Baker, Commented Oct 27, 2014 at 19:03
What is your question, friend?

typeoneerror
– typeoneerror

2014-10-27 19:03:40 +00:00
Commented Oct 27, 2014 at 19:03 — typeoneerror
– typeoneerror, Commented Oct 27, 2014 at 19:03
My question is why isn't the code above working i guess

Tim Mcneal
– Tim Mcneal

2014-10-27 21:49:04 +00:00
Commented Oct 27, 2014 at 21:49 — Tim Mcneal
– Tim Mcneal, Commented Oct 27, 2014 at 21:49

Community · Accepted Answer · 2023-11-17 19:24:28Z

1

There doesn't appear to be a question here, but perhaps mysqli_fetch_object would be more useful to you in this case.

Also, watch your naming. You've called your class "person" and instantiated it as "Person". PHP is rather flexible about case, but this could lead to issues later on in your program.

Lastly, json_encode just encodes an object. If you want to display the result or push it to the browser for consumption, you'll need to echo it out:

header('Content-Type: application/json');
echo json_encode($people);

edited Nov 17, 2023 at 19:24

CommunityBot

11 silver badge

answered Oct 27, 2014 at 19:08

typeoneerror

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Comments

Andrei Durduc · Accepted Answer · 2015-09-10 13:09:41Z

0

Use "mysqli_fetch_assoc" instead of "mysqli_fetch_array" if you want to use an associative approach.

class Person 
{
   public $firstName = null;
   public $lastName  = null;
}

$people = array();
$result = mysqli_query($con,"SELECT * FROM `People`");

while($row = mysqli_fetch_assoc($result)) {
      $e = new Person();        

      $e->firstName = $row['firstName'];
      $e->lastName  = $row['lastName'];   
}

json_encode($people);

answered Sep 10, 2015 at 13:09

Andrei Durduc

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