Php code to get json data from mysql database

Question

Iam trying to create a JSON object out of my SQL data in PHP. How to do that? This is my approach, which does not work so far.

<?php
header("Access-Control-Allow-Origin: *");
header('Content-Type: text/html; charset=utf-8');
$dns  = "mysql:host=localhost;dbname=afreen";
$user = "root";
$pass = "";
try {
    $conn = new PDO($dns, $user, $pass);
    if (!$conn) {
        echo "Not connected";
    }
    $query = $conn->prepare('SELECT id, name, salary from afreen');
    $query->execute();
    $registros = "[";
    while ($result = $query->fetch()) {
        if ($registros != "[") {
            $registros .= ",";
        }
        $registros .= '{"id": "' . $result["id"] . '",';
        $registros .= '"name": "' . $result["name"] . '"}';
        $registros .= '"salary": "' . $result["salary"] . '"}';
    $registros .= "]";
    echo $registros;
} catch (Exception $e) {
    echo "Erro: " . $e->getMessage();
}
?>`

Ranjit · Accepted Answer · 2017-12-13 13:30:58Z

2

Why dont you try the json_encode() for this ? Why you try for unnecessary while loop.

$query = $conn->prepare('SELECT id, name, salary from afreen');

Then try the json_encode() to get the Data in Json format.

Sources - http://php.net/json_encode

edited Dec 13, 2017 at 13:30

answered Dec 13, 2017 at 13:21

Ranjit

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Tarang koradiya · Accepted Answer · 2017-12-13 13:28:55Z

0

<?php
header("Access-Control-Allow-Origin: *");
header('Content-type: application/json');
header('Content-Type: text/html; charset=utf-8');
$dns  = "mysql:host=localhost;dbname=afreen";
$user = "root";
$pass = "";
try {
    $conn = new PDO($dns, $user, $pass);
    if (!$conn) {
        echo "Not connected";
    }
    $query = $conn->prepare('SELECT id, name, salary from afreen');
    $query->execute();
    $registros= [];
    while ($result = $query->fetch()) {

       array_push($registros,array(
                'id' =>$result["id"],
                'title' =>$result["name"],
                'salary' =>$result["salary"]
            ));
   $output = json_encode(array("product"=>$registros));
   print_r($output);
} catch (Exception $e) {
    echo "Erro: " . $e->getMessage();
}
?>

edited Dec 13, 2017 at 13:28

answered Dec 13, 2017 at 13:22

Tarang koradiya

7025 silver badges8 bronze badges

3 Comments

John Conde Over a year ago

Code dumps do not make for good answers. You should explain how and why this solves their problem. I recommend reading, "How do I write a good answer?"

khanum Over a year ago

Parse error: syntax error, unexpected 'catch' (T_CATCH) in C:\xampp\htdocs\appmarket-api\listado.php on line 25

Tarang koradiya Over a year ago

try and catch remove and check output

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