data in correct json format - php

Question

I am working on a feature of my website where I need to trigger an ajax call on the click of button. I am receiving data in this format:
Object {
name: "abc",
category_id: "1"
}
I want a result like this:
{
"list" : [
{
"name" : "xyz",
"category_id" : "1"
}, {
"name" : "abc",
"category_id" : "11"
}
]
}
my server side code is:
<?php
header('Content-Type: application/json');
$conn = mysqli_connect("localhost","root","","test") or die(mysqli_error());
$name = $_GET['name'];
$sql = "SELECT * from test_search where name = '$name'";
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
while($row = mysqli_fetch_array($result)) {
$name = $row['name'];
$category = $row['category_id'];
}
$data = array("name" => $name , "category_id" => $category );
echo json_encode($data);
?>

Answer 1

Before, you were assigning $row array items to variables, and then not using them until after the loop. After the loop, you used those variables, but it would only have the last values. You must append your items to an array ('list') inside of your array and then json_encode() your result from the loop to get the desired effect.
<?php
$conn = mysqli_connect("localhost","root","","test") or die(mysqli_error());
$name = $_GET['name'];
header('Content-Type: application/json');
$sql = "SELECT * from test_search where name = '$name'";
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
$data_array = array('list' => array());
while($row = mysqli_fetch_array($result))
{
$data_array['list'][] = array("name" => $row['name'], "category_id" => $row['category_id'] );
}
echo json_encode($data_array);
?>
Untested code, apologies for any syntax errors.
I would also like to point out that your code is open to SQL injection attacks. Follow the link to the PHP Manual to learn more about these and how to prevent them.

Answer 2

You need to push $name and $category variables into another array - currently you're simply overriding those values in your while loop.
Do something like:
$list = array();
while($row = mysqli_fetch_array($result))
{
$list[] = array('name'=>$row['name'], 'category_id'=>$row['category_id']);
}
$data = array('list'=>$list);
echo json_encode($data);

Answer 3

Have a read about PHP arrays here - http://php.net/manual/en/function.array.php
Also before you jump straight to json_encode($data), try var_dump($data) see what you're working with.
You need to build the array in your while loop,
$data = [];
while($row = mysqli_fetch_array($result)) {
$data['list'][] = [
'name' => $row['name'],
'category' => $row['category']
]
}
echo json_encode($data);
Untested, but should give you what you need.