If url contains - echo code

PHP
1

Hi guys,

How can perform the following:

if url conains:
categoryid=117
categoryid=123
categoryid=124
categoryid=125
categoryid=126

echo “clothing logo”;

if url conains:
categoryid=118

echo “play logo”;

if url conains:
categoryid=119

echo “home logo”;

$currentpage = $_SERVER[‘REQUEST_URI’];
…

2

You could use a series of “if … else” blocks to compare the supplied ID in turn and echo the appropriate reply.

3

thanks for the reply…
but what function do I use to check if url contains… ?

4

$categoryid = parse_str(parse_url($_SERVER[‘REQUEST_URI’], PHP_URL_QUERY));
$categoryid = $categoryid[‘categoryid’];

5

Why not fetch it via the GET array:



if (isset($_GET['categoryid'])) {
$categoryid=$_GET['categoryid'];
}
6

SpacePhoenix,

how do you do multiple checks at once (to limit coding)

if($categoryid==‘117’) or 118 or 119 {
do something
}

7

Wow, what was I thinking?! I was stupid. :rofl:

8

OK, I got this far…

Is this coding ok?


if (isset($_GET['categoryid'])) {
$categoryid=$_GET['categoryid'];

	if($categoryid=='117') {
	echo "clothing logo";
	} 

	elseif($categoryid=='123') {
	echo "clothing logo";
	} 
	
	elseif($categoryid=='124') {
	echo "clothing logo";
	}

	elseif($categoryid=='125') {
	echo "clothing logo";
	}

	elseif($categoryid=='126') {
	echo "clothing logo";
	}

elseif($categoryid=='118') {
	echo "playtime logo";
	}

elseif($categoryid=='119') {
	echo "home logo";
	}
	
elseif($categoryid=='120') {
	echo "baby logo";
	}


else	{
	echo '<div id="logo-pg-2"></div>';
}
9

damn… my if else structure must not be right… I got a blank page when trying to oad up the page with the code on it :frowning:

Can anyone possibly help?

10 
if (isset($_GET['categoryid'])) { 
$categoryid=$_GET['categoryid'];
} 

    if($categoryid=='117') { 
    echo "clothing logo"; 
    }  

    elseif($categoryid=='123') { 
    echo "clothing logo"; 
    }  
     
    elseif($categoryid=='124') { 
    echo "clothing logo"; 
    } 

    elseif($categoryid=='125') { 
    echo "clothing logo"; 
    } 

    elseif($categoryid=='126') { 
    echo "clothing logo"; 
    } 

elseif($categoryid=='118') { 
    echo "playtime logo"; 
    } 

elseif($categoryid=='119') { 
    echo "home logo"; 
    } 
     
elseif($categoryid=='120') { 
    echo "baby logo"; 
    } 


else    { 
    echo '<div id="logo-pg-2"></div>'; 
}

You would have had an error due to missing a } after the isset… block. To see the errors you will need to enable error display in php.ini

11

Or perhaps even an associated array:


$logoTypes = array(
    '117' => 'clothing',
    '123' => 'clothing',
    '124' => 'clothing',
    '125' => 'clothing',
    '126' => 'clothing',
    '118' => 'playtime',
    '119' => 'baby',
    '120' => 'baby'
);
echo $logoTypes[$categoryId] . ' logo';
12

i got it using



<?


$categoryid=$_GET['categoryid'];

	if($categoryid=='117') {
	echo '<div id="logo-clothing"></div>';
	} 

	elseif($categoryid=='123') {
	echo '<div id="logo-clothing"></div>';
	} 
	
	elseif($categoryid=='124') {
	echo '<div id="logo-clothing"></div>';
	}

	elseif($categoryid=='125') {
	echo '<div id="logo-clothing"></div>';
	}

	elseif($categoryid=='126') {
	echo '<div id="logo-clothing"></div>';
	}

elseif($categoryid=='118') {
	echo '<div id="logo-playtime"></div>';
	}

elseif($categoryid=='119') {
	echo '<div id="logo-home"></div>';
	}
	
elseif($categoryid=='120') {
	echo '<div id="logo-baby"></div>';
	}


else	{
	echo '<div id="logo-pg-2"></div>';
}


?>

thank yoou soooo much for your help

13

Or even, a structure based on the logo types:


$logoTypes = array(
    'clothing' => array('117', '123', '124', '125', '126'),
    'playtime' => array('118'),
    'baby' => array('119', '120')
);
foreach ($logoTypes as $logoType => $ids) {
    if (in_array($categoryId, $ids)) {
        echo $logoType . ' logo'; 
    }
}