Transfering arrays from php with json to browser - php

Question

I've got a php file containing.
<?php
$con=mysqli_connect("localhost", "user", "password", "stock");
if (mysqli_connect_errno()){
echo "failed to connect:" mysqli_connect_error();
}
$grabCars = mysqli_query($con, "SELECT * FROM CARS");
while ($row = mysqli_fetch_array($grabCars)){
$name = $row["Name"];
$color = $row["Color"];
$link = $row["Link"];
};
echo json_encode($name);
?>
ok can anyone tell me if there is anything wrong with this code. Any ideas on how this data would be displayed.
I could also do with some help at the other end what sort of jquery could would I use to read this data and how would it look, I'm very new to web design and don't know much jquery or how the ajax command would deal with this information.
Edit:
Current Jquery script
$.ajax({
url: "test.php"'
type: "post",
data: data,
datatype: "json",
success: function(result){
console.log(result["$name"]);
},
error: function(){
alert("error");
}
});
This is the code I've got to display some of the information in console, but I get nothing back, I get a data undefined message in console. Could really do with the help. Very new to json and jquery and php and webdesign as a whole. Thanks.

Answer 1

First of all, there shouldn't be ; after while loop.
$sampleArray = array();
while ($row = mysqli_fetch_array($grabCars)){
$name = $row["Name"];
$color = $row["Color"];
$link = $row["Link"];
array_push($sampleArray, array('name'=> $name, 'color' => $color, 'link'=>$link));
}
echo json_encode($sampleArray);
In your JQUERY/Javascript, something like this, in AJAX success:
response( $.map( data, function( item ) {
return {
name: item.name,
color: item.color,
link: item.link
}
}));

Answer 2

json_encode take one value parameter if you wanna encode multiple value you need to put them in an array
There is an example with your code
<?php
$con=mysqli_connect("localhost", "user", "password", "stock");
if (mysqli_connect_errno()){
echo "failed to connect:" mysqli_connect_error();
}
$grabCars = mysqli_query($con, "SELECT * FROM CARS");
$result = array();
$i = 0;
while ($row = mysqli_fetch_array($grabCars)){
$result[$i]['name'] = $row["Name"];
$result[$i]['color'] = $row["Color"];
$result[$i]['link'] = $row["Link"];
$i++;
}
echo json_encode($result);
?>