Search MYSQL database

[donthedragon]

Hi.

 

I want to implement a search on my website.

I have various venues with certain specifics on the site. eg : how many people it can accommodate, its location, style etc.

 

On my search, if someone puts in that they want the venue to accommodate 30 guests,

How can I get it to search for all venues that can accommodate up to 30 guests, but no more.

So all venues that can hold from 0 - 30 guests.

 

Kind regards

Donovan

[BLaZuRE]

Use

WHERE `your_field` < 30

 

Are you saying search as in text box or search as in searching your database with data provided from pre-made forms (drop-downs, checkboxes, etc)?

[donthedragon]

Oh brilliant. I'll definitely give that a try.

 

It is for a drop down in a search.

[donthedragon]

It seems to be working. Thanks.

 

I have some pagination that works with the search.

I only want it to show if the user needs to view more than 10 at a time.

If there is less than 10 I dont want the pagination to show at all.

Please can you help rectify my code below :

 

(I dont have a problem with this first code. I have just added it to show how it relates to the bottom code)

<?php
			 $area = $_POST['area'];
			 $max_guests_to = str_replace(" ","",$_POST['max_guests_to']);

			 $db = mysql_connect(localhost,'****','*****') or die("could not connect");
				mysql_select_db('******') or die( "Unable to select database");

			$query = "select * from property_profile where area='".mysql_real_escape_string($area)."' and max_guests < '".mysql_real_escape_string($max_guests_to)."'";
			$numresults = mysql_query($query) or die("Query error: ". mysql_error());
			$num_rows = mysql_num_rows($numresults);
			$page = 1;
			$prev_page = $page - 1;
			$next_page = $page +1;
			$num_pages = ceil($num_rows / 10) -1;
			$page_start = (10 * $page) - 10;
			$query = "select * from property_profile where area='".mysql_real_escape_string($area)."' and max_guests < '".mysql_real_escape_string($max_guests_to)."' LIMIT $page_start, 10";
			$result = mysql_query($query) or die("Query error: ". mysql_error());
			while($row = mysql_fetch_array($result)) {
				$reference = $row["reference"];
				$venue_name = $row["venue_name"]; 
				$area = $row["area"];
				$max_guests = $row["max_guests"];
				$description = $row["description"];
				$list_image = $row["list_image"];
				$star_grading = $row["star_grading"];
		?>

 

Below is the pagination code :

 

<?php
                	// Set up the page links and pass any needed variables to the next page by including them in the href after the page variable.
				if (!$prev_page < "1") {
					print "<p style='clear:both;'><a href='results.php?page=$prev_page&num_pages=$num_pages'>< Previous page.</p>";
				}
				if ($page != $num_pages) {
					print "<p style='clear:both;'><a href='results.php?page=$next_page&num_pages=$num_pages'>   Next page.></p>";
				}
			 ?>

[QuickOldCar]

I would think this should work. Give it a try.

 

<?php
                   // Set up the page links and pass any needed variables to the next page by including them in the href after the page variable.
               if (!$prev_page < 1) {
                  print "<p style='clear:both;'><a href='results.php?page=$prev_page&num_pages=$num_pages'>< Previous page.</p>";
               }
               
               if ($page >= 1 && $num_pages >= 2) {
               if ($page != $num_pages) {
                  print "<p style='clear:both;'><a href='results.php?page=$next_page&num_pages=$num_pages'>   Next page.></p>";
               }
               }
             ?>

[QuickOldCar]

Maybe you are happy with your code, but if wanted something more I wrote up a pagination script that controls the mysql start and stop rows.

 

Also added +x amount of pages if available and a go to page text area as well.

 

Here's the code in the forums

http://www.phpfreaks.com/forums/php-coding-help/results-on-multiple-pages-help/msg1506457/#msg1506457

 

If want to test it, here's the live demo.

http://get.blogdns.com/dynaindex/paginate.php

 

Maybe can even use something from it for yours.

[donthedragon]

Someone else did the code for me, so I would be more than happy to try something better.

Thanks, Ill check it out and let you know.