how to display in json format using php - php

Question

using php database connection i want to display data in json format which data are fatched from database(MySql),but i can't displaying in json format. http://takeyourtime.16mb.com/fatchData.php
$con = mysqli_connect($host, $username, $pwd, $db) or die('Unable to connect');
if (mysqli_connect_error($con))
{
echo "Failed to Connect to Database ".mysqli_connect_error();
}
$name = $_POST['Query'];
$sql = "SELECT * FROM playerstb";
$query = mysqli_query($con,$sql);
if ($query)
{
$rows = array();
while ($r = mysql_fetch_assoc($query)) {
$rows['root_name'] = $r;
}
}
echo json_encode($rows);
mysqli_close($con);

Answer 1

Just use json_encode. BTW, your script has an syntax error in the ending if block:
if($query){
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$rows['root_name'][] = $r; // probably must be an array
}
echo json_encode($rows);
}else{
/*
This will show up when you have a query error
nothing to do with the results found.
I would consider changing the message below
*/
echo('Not Found');
}

Answer 2

inside your while loop you dont save all results you each one writen over the before one
you have to store it in array like note this ([])*
while($r = mysql_fetch_assoc($query)) {
$root_names[] = $r;
}
echo json_encode(['root_name'=>$root_names]);

Answer 3

You have to store first your result in array then after that create an array name your desire key ($array["name"])
$con=mysqli_connect($host,$username,$pwd,$db) or die('Unable to connect');
if(mysqli_connect_error($con))
{
echo "Failed to Connect to Database ".mysqli_connect_error();
}
$name=$_POST['Query'];
$sql="SELECT * FROM playerstb";
$query=mysqli_query($con,$sql);
if($query)
{
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$rows[] = $r;
}
$data["data"]=$rows;
echo json_encode($data);
}
}else
{
echo('Not Found ');
}
mysqli_close($con);
?>