Jquery json array from php

Question

I am trying to reproduce the following but my using rows from the database

var allCustomers = [

    { name: 'Customer1', contacts: [
                { name: 'Bob', id: ['1'] },
                { name: 'Sue', id: ['2'] },
                { name: 'John', id: ['3'] }
        ]},

    { name: 'Customer2', contacts: [
                { name: 'Max', id: ['4'] },
                { name: 'Ross', id: ['5'] },
                { name: 'Sally', id: ['6'] }
        ]}
];

In PHP I am fetching the rows from the database, each customer has multiple contacts, which is the bit I am struggling with. Currently I am using the method below:

<script type="text/javascript">
var allCustomers = [
<?php
include('connection.php');
$stmt = $db->prepare("SELECT customer.customerID, customerName, contactID, contactName  FROM customer INNER JOIN customerContact ON customer.customerID = customerContact.customerID");
if ($stmt->execute()) 
{
    while ($row = $stmt->fetch(PDO::FETCH_OBJ)) 
    {
            echo "{ name: '".$row->customerName."', contacts: [
                { name: '".$row->contactName."', id: ['".$row->contactID."'] }                
        ]},";
        }           
}       
?>    
];

But obviously this isn't a very neat way of doing this, and it only works if a customer has only one contact, otherwise it reproduces the customer and contact again.

What would you suggest to fix this, could I make use of the php decode json function or something similar?

Thanks for any suggestions :).

I am trying to produce something similar to this post, but I need the id of the contact to post back to the server.

Imi Borbas · Accepted Answer · 2011-03-28 21:50:56Z

2

Something like this:

$data = array(
  array("name" => "Customer1", "contacts" => array(
    array("name" => "Bob", "id" => 1),
    array("name" => "Sue", "id" => 2),
    array("name" => "John", "id" => 3)
  )),
  array("name" => "Customer1", "contacts" => array(
    array("name" => "Max", "id" => 4),
    array("name" => "Ross", "id" => 5),
    array("name" => "Sally", "id" => 6)
  ))
);

echo json_encode($data);

In reply to the asker's comment:

$stmt = $db->prepare("SELECT customer.customerID, customerName, contactID, contactName  FROM customer INNER JOIN customerContact ON customer.customerID = customerContact.customerID");

$data = array();

if ($stmt->execute()) 
{
    while ($row = $stmt->fetch(PDO::FETCH_OBJ)) 
    {
        if (!isset($data[$row->customerId])
        {
            $data[$row->customerId] = array('name' => $row->customerName, 'contacts' => array());
        }

        $data[$row->customerId]['contacts'][] = array('name' => $row->contactName, 'id' => $row->contactId);
    }           
}   

echo json_encode(array_values($data));

edited Mar 28, 2011 at 21:50

answered Mar 28, 2011 at 20:48

Imi Borbas

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3 Comments

Elliott Over a year ago

Thanks, I did originally do it this way but am abit confused to how I would loop through the multiple contacts for each customer?

Elliott Over a year ago

Thanks alot that exactly how I wanted it :) Just one more thing is it possible to not include the customerid when I echo the array back out? Thanks again.

Imi Borbas Over a year ago

Updated the last row. Hope this way it'll work as you expect.

remi.gaubert · Accepted Answer · 2011-03-28 20:47:32Z

0

For create JSON with PHP, you can see this tutorial : http://www.sencha.com/learn/Tutorial:Creating_JSON_Data_in_PHP

answered Mar 28, 2011 at 20:47

community wiki

remi.gaubert

Comments

Yoram de Langen · Accepted Answer · 2011-03-28 20:49:01Z

0

Does do everything:

echo json_encode($array);

answered Mar 28, 2011 at 20:49

Yoram de Langen

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