how to call a class function from jquery php? - php

Question

I'm working with the plugin ParamQuery grid and to the data source need to get json data,
this what I want to do with $. getJson (...) {}, but I have a PHP class called data.php, which contains
a method called GetData other InsertData, DeleteData, (CRUD-using PDO), GetData returns the infromacion as json.
The problem is how to call the function from jquery?
the code I use is:
data.php
<?php
class Data {
private $db = NULL;
const DB_SERVER = "localhost";
const DB_USER = "root";
const DB_PASSWORD = "usbw";
const DB_NAME = "musicstore";
public function __construct() {
$dsn = 'mysql:dbname=' . self::DB_NAME . ';host=' . self::DB_SERVER;
try {
$this->db = new PDO($dsn, self::DB_USER, self::DB_PASSWORD);
} catch (PDOException $e) {
throw new Exception('Connection failed: ' . $e->getMessage());
}
return $this->db;
}
public function getData() {
$statement = $this->db->prepare("Select * from Customer");
$statement->execute();
if ($statement->rowCount() > 0) {
echo json_encode($statement);
}
return false;
}
}
?>
functionjquery.js
$.getJSON('Data.php', function(data) {
var obj = {};
obj.width = 1000;
obj.height = 400;
obj.colModel = [{title: "Rank", width: 150, dataType: "integer"},
{title: "Company", width: 200, dataType: "string"},
{title: "Revenues ($ millions)", width: 200, dataType: "float", align: "right"},
{title: "Profits ($ millions)", width: 200, dataType: "float", align: "right"}];
obj.dataModel = {data: data};
$("#grid_array").pqGrid(obj);
});

Answer 1

You need to create a page which constructs an instance of the Data class and then outputs the results of getData()
You shouldn't be echoing from inside a function though. Change your getData method to something like the following:
public function getData() {
$statement = $this->db->prepare("Select * from Customer");
$statement->execute();
return $statement->rowcount() > 0 ? $statement->fetchAll() : NULL;
}
Then, create a new page let's call it json_data.php for transparency's sake:
require_once "data.php";
$dataObj = new Data();
echo json_encode($dataObj->getData());
Now change your $.getJSON call in jQuery to request the json_data.php page.
$.getJSON('json_data.php', function(data) { ... });
And that should work.

Answer 2

Actually you can pass the parameter data along with the $.getJSON request to identify which method to be executed.
For Eg.
in functionjquery.js
$.getJSON('data.php', { method:'get' }, function(data) { ... });
in data.php
Change it as #Jason stated for best practice, and within that
if(isset($_GET['method'])){
switch (($_GET['method'])) {
case 'get':
echo json_encode(getData());
exit();
//other cases go here
default:
break;
}
}
In this way no need to create additional pages for each method