CODEWITHSUNDEEP
phpjquerymysqlajax
Steven Sugiarto Wijaya
Steven Sugiarto Wijaya

Reputation: 171

double live search using ajax

following from this sample: https://codeforgeek.com/2014/09/ajax-search-box-php-mysql/

i create one input livesearch. but i want more input. this is what have i do:

$(document).ready(function(){
    $('input.nama').typeahead({
        name: 'nama',
        remote:'search.php?keynama=%QUERY',
        limit : 10
    });

     $('input.cabang').typeahead({
        name: 'cabang',
        remote:'search.php?keycabang=%QUERY',
        limit : 10
    });

});

this is input type:

<input type="text" name="nama" class="nama livesearch tt-query" autocomplete="off" spellcheck="false" placeholder="Nama">
<input type="text" name="cabang" class="cabang livesearch tt-query" autocomplete="off" spellcheck="false" placeholder="Cabang">

the last code in search.php

$keynama=$_GET['keynama'];
    $nama = array();

    //$keycabang=$_GET['keycabang'];
    //$cabang = array();

    $con=mysql_connect("localhost","root","");
    $db=mysql_select_db("dbdataorder",$con);

    if($keynama!=null)
    {
        $query=mysql_query("SELECT  idNoon,notaris,noon,cabang,nama,proses,tanggalOn,tanggalCair,keterangan 
        FROM tblNotaris,tblOrderNotaris
        where tblNotaris.idnotaris=tblOrderNotaris.idnotaris and nama LIKE '%{$keynama}%'");

        while($row=mysql_fetch_assoc($query))
        {$nama[] = $row['nama'];}
        echo json_encode($nama);
    }

when declare get value key cabang the first livesearch "name" became not work. i think there is something wrong with my declaration. maybe when passing value from index.php to search.php data not send. or an error in jquery. any help would be appreciated. thanks

Upvotes: 1

Views: 120

Answers (1)

himeshc_IB
himeshc_IB

Reputation: 863

It is working properly... you need to update your search.php if you want to make second text box work....

$keynama=$_GET['keynama'];
$nama = array();

$keycabang=$_GET['keycabang'];
$cabang = array();

$con=mysql_connect("localhost","root","root");
$db=mysql_select_db("dbdataorder",$con);

if($keynama!=null)
{
    $query=mysql_query("SELECT  idNoon,notaris,noon,cabang,nama,proses,tanggalOn,tanggalCair,keterangan 
    FROM tblNotaris,tblOrderNotaris
    where tblNotaris.idnotaris=tblOrderNotaris.idnotaris and nama LIKE '%{$keynama}%'");

    while($row=mysql_fetch_assoc($query))
    {$nama[] = $row['nama'];}
    echo json_encode($nama);
}


if($keycabang!=null)
{
    $query=mysql_query("SELECT  idNoon,notaris,noon,cabang,nama,proses,tanggalOn,tanggalCair,keterangan 
    FROM tblNotaris,tblOrderNotaris
    where tblNotaris.idnotaris=tblOrderNotaris.idnotaris and cabang LIKE '%{$keycabang}%'");

    while($row=mysql_fetch_assoc($query))
    {$cabang[] = $row['cabang'];}
    echo json_encode($cabang);
}

And JS code is given below...

$(document).ready(function(){

    $('input.nama').typeahead({
        name: 'nama',
        remote:'search.php?keynama=%QUERY',
        limit : 10
    });

     $('input.cabang').typeahead({
        name: 'cabang',
        remote:'search.php?keycabang=%QUERY',
        limit : 10
    });

});

Upvotes: 1

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