PHP filtering results

[usman07]

I want to create a feature as you see nn this image, where it says 'filter your results' if a user clicks 'Detached Houses' then only detached houses will be displayed. if a user clicks 'Semi-detached' then only semi detached houses will be shown.

Any help is really appreciated, thank you.

<table border="0" cellpadding="0" cellspacing="0">
<tr>
<td><div id="filter"><p class="houses" style="font-family:helvetica;color:#0155a1;font-size:14px;background:url(cutouts/forsale/filter.jpg) no-repeat;"><b><u>Houses</u></b>
					<br />
					<span class="dh"><b><u>Detached Houses</u></b></span><?php
// Make a MySQL Connection
mysql_connect("localhost", "admin", "1admin") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

// Get a specific result from the "example" table
$result = mysql_query("SELECT * FROM example
WHERE name='Sandy Smith'") or die(mysql_error());  

// get the first (and hopefully only) entry from the result
$row = mysql_fetch_array( $result );
// Print out the contents of each row into a table 
echo $row['name']." - ".$row['age'];
?>
					<br />
					<span class="dh"><b><u>Semi-detached houses</u></b></span>
					<br />
					<span class="dh"><b><u>Terraced houses</u></b></span>
					<br />
					<br />
					<b><u>Flats / Apartments</u></b>

</p></div></td>

[Jessica]

How many topics have you created about this now?

 

Make the text a link, and in the link add a parameter of what you want to search on.

It should look like: http://mysite.com/show_houses.php?type=semi-detached

 

Then on your page you'll check if $_GET['type'] is set and has value, and then create your MySQL query to use that.

 

You have to try it before we can help you fix it.

[usman07]

Im sorry I don't mean to misuse the forums. Thanks for that information, really appreciate it. i will try it.

[usman07]

Is this looking ok so far?

<div id="main">

<table border="0" cellpadding="0" cellspacing="0">
<tr>
<td><div id="filter"><p class="houses" style="font-family:helvetica;color:#0155a1;font-size:14px;background:url(cutouts/forsale/filter.jpg) no-repeat;"><b><u>Houses</u></b>
					<br />
	<a href="www.mumtazproperties.hostei.com/show_houses.php?type=semi-detached"><span class="dh"><b><u>Detached Houses</u></b></span></a>
<?php
// Make a MySQL Connection
mysql_connect("", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());

// Get a specific result from the "example" table
$result = mysql_query("SELECT * FROM example
WHERE name='Sandy Smith'") or die(mysql_error());  

// get the first (and hopefully only) entry from the result
$row = mysql_fetch_array( $result );
// Print out the contents of each row into a table 
echo $row['name']." - ".$row['age'];
?>
					<br />
					<span class="dh"><b><u>Semi-detached houses</u></b></span>
					<br />
					<span class="dh"><b><u>Terraced houses</u></b></span>
					<br />
					<br />
					<b><u>Flats / Apartments</u></b>

 

One thing u mentioned that im not sure about is 'Then on your page you'll check if $_GET['type'] is set and has value, and then create your MySQL query to use that.'

I have limited knowledge with PHP code i'm afraid, but do i need to set a GET type in the php code?

 

Thank You

 

[Psycho]

One thing u mentioned that im not sure about is 'Then on your page you'll check if $_GET['type'] is set and has value, and then create your MySQL query to use that.'

I have limited knowledge with PHP code i'm afraid, but do i need to set a GET type in the php code?

 

Values passed on the URL are available in the global $_GET array. So, if the user access a page using

http://mysite.com/page.php?foo=bar&this=that

 

Then when that script runs the $_GET array will look like this

array(
    'foo' => 'bar',
    'this' = 'that'
)

 

So, to expand on what jesirose suggested, you would do something like this:

$query = "SELECT * FROM example";
if(isset($_GET['type']))
{
    $query .= " WHERE type = '" . mysql_real_escape_string($_GET['type']) . "'";
}

[usman07]

Is this the correct way to do this?

<table border="0" cellpadding="0" cellspacing="0">
<tr>
<td><div id="filter"><p class="houses" style="font-family:helvetica;color:#0155a1;font-size:14px;background:url(cutouts/forsale/filter.jpg) no-repeat;"><b><u>Houses</u></b>
					<br />
	<a href="www.mumtazproperties.hostei.com/show_houses.php?type=detached"><span class="dh"><b><u>Detached Houses</u></b></span></a>
<?php
// Make a MySQL Connection
mysql_connect("", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());

// Get a specific result from the "example" table
$result = mysql_query("SELECT * FROM example
WHERE name='Sandy Smith'") or die(mysql_error());

$query = "SELECT * FROM example";
if(isset($_GET['type']))
{
    $query .= " WHERE type = '" . mysql_real_escape_string($_GET['type']) . "'";
}  

// get the first (and hopefully only) entry from the result
$row = mysql_fetch_array( $result );
// Print out the contents of each row into a table 
echo $row['name']." - ".$row['age'];
?>
					<br />
					<span class="dh"><b><u>Semi-detached houses</u></b></span>						
					<br />
					<span class="dh"><b><u>Terraced houses</u></b></span>
					<br />
					<br />
					<b><u>Flats / Apartments</u></b>

</p></div></td>

 

where it says 'example' in the php code, is that where I put the name of my table in MYSQL?

Basically what i want it to do is when a user clicks on "detached houses", all the detached house will be displayed from the database.

 

Thank you for your help, much appreciated.

[usman07]

I desperately need some help, any help would be hugely appreciated.

[Zephni]

<?php
  $query = mysql_query("SELECT * FROM something WHERE name='someone'");
  $result = mysql_fetch_array($query);

  var_dump($result);
?>

 

Get the result of the query as an array with mysql_fetch_array(); Then access the items with $result['id'] or $result['surname']

[usman07]

Thanks for your reply Zephni, Where abouts would I placed that code in within my previous post of PHP coding? Thanks