i'm trying create php mysql search engine working e-comerce search engine auto suggestion using ajax?... table like
id cat name 1 men subi 2 men flick 3 women sheeba 4 women leena my form
<html> <head> <title>search engine</title> </head> <body> <form action = 'ss.php' method ='get'> <input type = "text" name = "q"> <input type = "submit" name = "submit" value = "search" </body> </html> and ss.php is
$k = $_get["q"]; $con = mysqli_connect("localhost", "root", ""); mysqli_select_db($con,"x"); $terms=explode(" ",$k); $i=0; $set_limit = ("9"); $subi = ""; foreach ($terms $each) { $i++; $escapedsearchstring = mysqli_real_escape_string($con,$each); if ($i == 1 ) $subi.= " title '%$escapedsearchstring%' "; else $subi.= " , title '%$escapedsearchstring%' "; } $query = "select sql_calc_found_rows * table $subi order rand() limit $set_limit"; $qry = mysqli_query($con,"$query"); $row_object = mysqli_query($con,"select found_rows() rowcount"); $row_object = mysqli_fetch_object($row_object); $actual_row_count = $row_object->rowcount; $result = $actual_row_count; it's work fine when search words subi or sheeba bt want if i'm start type word 's' 'll shows auto suggestion like
sheeba subi sheeba in women subi in men if user clicks sheeba query automaticaly changed
" select * table title '%sheeba%' " and if user clicks 'sheeba in women' 'll query changed
" select * table cat = 'women' , title '%sheeba%' " how can obtain this? pls answer briefly... tnx in advance....
if want create custom code rather using plugin need
$(document).ready(function() { $("input[name='q']").on("keyup",function(event){ search_value = $(this).val(); // check whether input not empty or has characters if(value.length > 0){ $.ajax({ url: '/path/to/file', type: 'get', datatype: 'json', data: "q="+search_value, success: function(response){ // create suitable html body show reponse $("suggestion_box_id").html("your/created/htmlcontent"); } }) } else { $("suggestion_box_id").html(""); } }) }); 
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