jQuery autocomplete from php-json without database

Question

I want to make autocomplete on my form with jQuery. The source taken from json data from php, but i don't use database. Here is my code :

facebookfriends.php

<?php
include('facebookdata.php');

$user_friend = $user_friends['data'];

$json_friends = json_encode($user_friend);
echo $json_friends;
?>

Script

$(function() {
  $( "#search" ).autocomplete(
  {
     source:'facebookfriends.php',
  });
});

JSON DATA

[{"name":"Indiarto Priadi","id":"502163984"},
 {"name":"Agustin Kertawijaya","id":"511990261"},
 {"name":"Jecklyne Christin L","id":"528197912"},
 {"name":"Jazi Eko Istiyanto","id":"531149275"},
 {"name":"Esritha Situmorang","id":"535864892"},
 {"name":"Christy Margaretha Siregar","id":"543468540"},
 {"name":"Daniel Oscar Baskoro","id":"549332828"},
 ........]

I just want to display the name in autocomplete to ensure that the autocomplete works well. But it doesn't. Please help. Thank you!

the autocomplete does not appear (nothing happens when I type some letter in input fields) — King Goeks
– King Goeks, Commented Nov 29, 2013 at 4:59
Autocomplete needs either an array of strings (without keys), or an array with keys of label and/or value - api.jqueryui.com/autocomplete/#option-source. Can you update facebookdata.php to add a label key, which could just be a copy of your name value. — Sean
– Sean, Commented Nov 29, 2013 at 5:07

AAA · Accepted Answer · 2013-11-29 05:54:09Z

1

1.You have to parse json in php 2.make an array for dropdown list

Following steps to be followed:

     $data ='[{"name":"Indiarto Priadi","id":"502163984"},
              {"name":"Agustin Kertawijaya","id":"511990261"},
              {"name":"Jecklyne Christin L","id":"528197912"}, 
              {"name":"Jazi Eko Istiyanto","id":"531149275"},
              {"name":"Esritha Situmorang","id":"535864892"},
              {"name":"Christy Margaretha Siregar","id":"543468540"},
              {"name":"Daniel Oscar Baskoro","id":"549332828"}]';
    $user_friend =  json_decode($data, true );
    $data=array();
    foreach($user_friend as $key=>$val)
            $data[]=$val['name'];
    $json_friends =json_encode($data);
    echo $json_friends;

answered Nov 29, 2013 at 5:54

AAA

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Comments

Padmanathan J · Accepted Answer · 2013-11-29 05:13:08Z

0

you need to pass label and value into you php file

For example

    $data[] = array(
                'label' =>  $row['city_name'].','.$state['state_name'].','.$c_name['country_name'],
                'value' =>  $row['city_name'].','.$state['state_name'].','.$c_name['country_name']
                    );

echo json_encode($data);
flush();

answered Nov 29, 2013 at 5:13

Padmanathan J

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