A better way to generate this json array from MySql data with php

Question

Sql fiddle for your convenience here.

I\'m taking data from a MySql table and turning it into a json array. All works well and I have the output the way I want it, but i

Answer 1

Taking out all the redundancy, using proper prepared statements (assuming PDO) and adding error handling (at least a stub), you end up with this:

$stmt = $conn->prepare('SELECT name, age, address, pincode FROM json WHERE name = ?');
$stmt->execute(array('peter'));

if ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
    echo json_encode($row);
} else {
    echo json_encode(array('status' => 'error'));
}

If you expect multiple rows:

echo json_encode($stmt->fetchAll(PDO::FETCH_ASSOC));

Answer 2

$row = $stmt->fetch(PDO::FETCH_ASSOC);
echo json_encode($row);

You don't need the while loop, if you only have one row in your result.

Answer 3

You only select needed fields, so you can just do

echo json_encode($stmt->fetch(PDO::FETCH_ASSOC));

EDIT: You mention now that name is unique, so above is all you need.