Auto complete multiple results in textfield from mysql database [closed] - php

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I am new in php and ajax. I have search a lot from internet for this problem but in vain. I want to make an auto complete system just like this website: (http://www.zameen.com/). My problem is more simple than this website. My requirement is: "When I will enter something in text field, it will show the match results from database. When i will select one of those results then it will show in the same text field. I want to select multiple results and all results will show in the same text field."
Now my next question is: "how i will handle all these results in php variables and post them to my required file."
I am so much confused. Please give a simple solution because I am new in php and ajax.
You can see an image to understand my problem on this link: (http://mutaliapakistan.com/problem.png).

Answer 1

I'll try to give you a simple, working example of what you want:
this will be your livesearch.js file:
function showResult(str)
{
if (str.length==0)
{
document.getElementById("livesearch").innerHTML="";
document.getElementById("livesearch").style.border="0px";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("livesearch").innerHTML=xmlhttp.responseText;
document.getElementById("livesearch").style.border="1px solid #A5ACB2";
}
}
xmlhttp.open("GET","livesearch.php?q="+str,true);
xmlhttp.send();
}
And this will be your livesearch.php file:
<?php
$hint="";
//get the q parameter from URL
if (!isset($key))$key = $_GET['q'];
//include the mysql conn script
require_once ('mysql_connect.php');
//just change the selects here to what you want/ need.
$query = "SELECT DISTINCT * FROM table_name WHERE display='1' AND (name LIKE '%$key%' OR description LIKE '%$key%') LIMIT 5";
$result = mysql_query($query);
while($result_row = mysql_fetch_array($result))
{
$found = $result_row[12];
if (strlen($found)>25)
{
$found = substr($found, 0, 25) . '...';
}
$hint = $hint."<div id='search_result'>".$found."</div>";
}
// Set output to "no suggestion" if no hint were found
// or to the correct values
if ($hint == "")
{
$response = "No results";
}
else
{
$response = $hint;
}
//output the response
echo $response;
?>
And now finally to the page that you want to have the seach in it add the following:
<script type="text/javascript" src="livesearch.js"></script>
<div class="search">
<form name="frm_search" id="frm_search" method="get" action="advanced_search.php">
<input type="text" name="search" id="search" value="" placeholder=" Search" onkeyup="showResult(this.value)"/>
<input type="submit" name="search_go" id="search_go" value="↵"/>
</form>
<div id="livesearch" style="margin-left:3px;background-color:#fff;text-align:left;">
</div>
</div>
I hope it's clear.