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05.18.2008 at 08:09AM PDT, ID: 23411926
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7.8

Building php search engine, with relevant results

Asked by socross in PHP and Databases, MySQL Server, SQL Query Syntax

Tags:

HI

I have been having real trouble getting my head around displaying relevant results, i have asked various questions with regads the mysql SELECT statement but have real issues working out the next bit.

I will attach untested code, which is using some tutorials and some of my own functions.

The bit i am struggling with is the ordering of results, at the moment i take the search string and splt it into an array, with each value being one of the words.

I then SELECT results from my database which match these words and store the row ids in a new array.

I use a case statement to order the SELECT results.

The bit im stuck on is ordering the results depending on how many of the original string keywords are found in a row!!

Here is my code

Any suggestions would be much appriciated!

--s--Start Free Trial 
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//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//--//
//
//--//--//  PROCESS SEARCH  //--//	
 
 
// Specify how many results to display per page
	$limit = 8;
// Get the search variable from URL
	$var = @$_POST['search-input'];
// Trim whitespace from the stored variable
	$trimmed = trim($var);  
// Separate key-phrases into keywords
	$trimmed_array = explode(" ",$trimmed);
// Check for an empty string and redirect || Check for no search var and redirect
	if ($trimmed == "" || !isset($var))
  		redirect('/search-empty/');
		
		
// Build SQL Query for each keyword entered 
foreach ($trimmed_array as $trimm){
 
	// SET QUERY VARS
	$db_table = 'for_search';
	$db_values = array('mach_id');
	$db_where = 'WHERE manu_name LIKE "%'.$trimm.'%" OR name LIKE "%'.$trimm.'%" OR details LIKE "%'.$trimm.'%"';
	$db_order = 'ORDER BY (CASE WHEN manu_name LIKE "%'.$trimm.'%" THEN 1 ELSE 0 END + CASE WHEN name LIKE "%'.$trimm.'%" THEN 1 ELSE 0 END + CASE WHEN details LIKE "%'.$trimm.'%" THEN 1 ELSE 0 END) DESC';
	$db_return = 'm_arr';
	
	//select_record($db_table, $db_values, $db_where, $db_order, $db_return);
	
      
 
     // Execute the query to  get number of rows that contain search kewords
     $row_num_links_main = get_rows($db_table, $db_values, $db_where, $db_order, $db_return);
 
     // next determine if 's' has been passed to script, if not use 0.
     // 's' is a variable that gets set as we navigate the search result pages.
     if (empty($s)) {
         $s=0;
     }
 
      // now let's get results.
     $db_order .= " LIMIT $s,$limit" ;
     $row_num_links_main = select_record($db_table, $db_values, $db_where, $db_order, $db_return);
 
      //store record id of every item that contains the keyword in the array we need to do this to avoid display of duplicate search result.
	  foreach($row_num_links_main as $r)
          $adid_array[] = $r['mach_id'];
	  }
 } //end foreach
 
if($row_num_links_main == 0 && $row_set_num == 0){
	
	
	// Put no search results redirects here - send get var with search
	redirect('/search-empty/');
 
}
   //delete duplicate record id's from the array. To do this we will use array_unique function
   $tmparr = array_unique($adid_array); 
   $i=0; 
   foreach ($tmparr as $v) { 
       $newarr[$i] = $v; 
       $i++; 
   }
 
// now you can display the results returned.
[+][-]05.18.2008 at 08:47AM PDT, ID: 21593076

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About this solution

Zones: PHP and Databases, MySQL Server, SQL Query Syntax
Tags: PHP MYSQL SQL
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Solution Provided By: DarkoLord
Participating Experts: 2
Solution Grade: A
 
 
[+][-]05.18.2008 at 02:57PM PDT, ID: 21594203

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[+][-]05.18.2008 at 04:27PM PDT, ID: 21594447

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[+][-]05.18.2008 at 04:35PM PDT, ID: 21594472

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[+][-]05.19.2008 at 06:19AM PDT, ID: 21597483

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[+][-]05.25.2008 at 03:14PM PDT, ID: 21643799

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