从mysql表中获取数据并将值放在JSONObject中

Question

I am fetching values from mysql table. I am then placing the values inside an array and using th php built function json_enconde to convert these values into a json format. I am getting results back but not in the desired format. In the php code you will notice I am using array and fetching values with a foreach loop to place all values inside. As a result I am getting back several json objects. How can I get back just one json object with the corresponding values?

header("Content-type: application/json");   
$query = $db_con->prepare("SELECT id, name FROM table");
$query->execute();
$data = $course_query->fetchAll();
$json_data = array();
foreach ($data as $row) {
    $json_data[] = array(
       $row["id"] => $row["name"]
    );
}  // foreach ($data as $row) {
echo json_encode($json_data);

Results:

{
1: "Item1"
},
{
2: "Item2"
},
{
3: "Item3"
}

Desired Results:

{
"1":"Item1",
"2":"Item2",
"3":"Item3"
}

Answer 1

try this

foreach ($data as $row) {
$id=$row["id"];
    $json_data[$id] =$row["name"];

}  

The problem was that while defining the value in json_data in your foreach loop you are creating an array for every single value, so you are not getting the desired result.