Tech World : Search suggestion using ajax,php and mysql

Create search engine with search suggestion like google using ajax,php and mysql.(search suggestion in php, how to show search suggestion using php and mysql,search. )

Step 1 : Create html text box (search.html )

<html>

         <title>search suggestion</title>
         <head>
<style type="text/css">
               #search_item ul li{display:block;}
        </style>
        <script type="text/javascript">
               $(document).ready(function(){
               $('#search_item ul').click(function(event){
                      var data = $(event.target).text();
                      $('#search_box').val(data);
                      $("#search_item").hide();
                  });
            });
</script>
         </head>

<body>

</form>
   <div class="search_item" id="search_item">
                    <ul>
                </ul>
                </div>

</body>

</html>

Step 2 : Create database table

CREATE TABLE IF NOT EXISTS `table1` (
`id` bigint(12) NOT NULL AUTO_INCREMENT,
`name` text NOT NULL,
PRIMARY KEY (`id`)
) ;

Step 3 : Create Ajax file ( ajax.js )

$(document).ready(function(){

$("#search_box").keyup(function(){
         var search=$("#search_box").val();
         var dataString="item="+search;
       $.ajax({
               type:"POST",
               url:"/ajax.php",
               data:dataString,
               cache: false,
                   success:function(response){
                   $("#rr ul").html(response);
                   },
                   fail:function(error){
                       alert(error);
                       }
           });
     });

});

Step 4 : Create php file ( ajax.php )

<?php

        $searchq=$_POST["item"];
        $keyword =preg_replace('/\s+/', '', $searchq);
        $keyword =str_split($keyword);

$query = "SELECT name FROM table1" . "WHERE name LIKE '%".$keyword['0']."%'";
              for ($i=1; $i<count($keyword); $i++) {
                    $query = $query."AND name LIKE '%".$keyword[$i]."%'";
              }

   $result = mysql_query($query);
                      while($row = mysql_fetch_assoc($result)) {
                              echo "<li>".$row['name']."</li>";
                         }

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Sunday, 28 April 2013

Search suggestion using ajax,php and mysql

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