MySQL Data to JSON Array with inserting a string

Question

This is my Code for now but it only works until the for loop in the mySQLResultsetToJSON().

<?php    
$servername = "127.0.0.1";
$username = "root";
$password = "123456";
$table = "ticketing";
$link = new mysqli($servername, $username, $password, $table);
if ($link->connect_error) {
    die("Connection failed: " . $link->connect_error);
}
    echo "Connected successfully";    

$result = $link->query("SELECT * from Ticket");

print_r($result);
$final_result = mySQLResultsetToJSON($result);
print_r($final_result);    
$link->close();    

function mySQLResultsetToJSON($resultSet)
{
        for($i = 0; sizeof($resultSet); $i++)
        {
            $rows = array();
            while($r = mysqli_fetch_assoc($resultSet[$i])) {
                $rows[] = $r;
            }
            $jsonResult[$i] = json_encode(array('Results' => $rows));
        }    
        print_r($jsonResult);    
        return $jsonResult;
    }    
?>

Thank you!

Thomas

no at the end i need an JSON array for my JS script to creat all the tickets. — Thomas Haas
– Thomas Haas, Commented May 4, 2016 at 5:44

Deepak Adhikari · Accepted Answer · 2016-05-04 05:50:56Z

1

echo "mysql data<br />";
$result = $link->query("SELECT * from users");
print_r($result->fetch_object());

echo "<br />";

echo "in json<br />";
$res = ['Results' => $result->fetch_object() ];
echo json_encode($res);
$link->close();

answered May 4, 2016 at 5:50

Deepak Adhikari

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1 Comment

Aman Sharma Over a year ago

Please consider adding details in your answer along with the code snippet, as to what you are doing and why you are doing it.

Brijal Savaliya · Accepted Answer · 2016-05-04 05:46:28Z

0

User like

$result = $link->query("SELECT * from Ticket");
$rows = array();
while($r = mysqli_fetch_assoc($result)) {
    $rows[] = $r;
}
print "<pre>";
print_r(json_encode(array('Results' =>$rows)));    
$link->close();

answered May 4, 2016 at 5:46

Brijal Savaliya

1,0919 silver badges19 bronze badges

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