if statement depending on file page name?

[Darkwoods]

Hey im having troube making this code as i have no idea how to do it...

 

basically what im trying to do is for example if the index.php?id=XXX i want to echo "this" and if it is just index.php i want to echo "that"

 

your help would be nice thanks

 

		<?php
	$id_lang = $_GET['id'];
	if($id_lang == $id_lang){
        $result = mysql_query("SELECT * FROM pages WHERE id='$id_lang' ",$connect);
        while($row44 = mysql_fetch_assoc($result))
		{
			$js_langdir = $row44['lang_dir']; }}

	else{
	$js_langdir = "LTR";
				}
?>

[fortnox007]

not sure if I understand your question fully but you might do this.

 

if (empty($_GET['id'])){
  echo 'this';  //this will be shown when someone types in  http://lalalaala.com/index.php?id=monkeys
}else{
  echo 'that';
}

[Darkwoods]

thank youu

 

this worked very well.. thanks again

[fortnox007]

no problemo, makes sure though this only checks if the $_GET['id'] is empty so if you are using other variables you might have to change something. but nothing difficult

[gevans]

not sure if I understand your question fully but you might do this.

 

if (empty($_GET['id'])){
  echo 'this';  //this will be shown when someone types in  [url=http://lalalaala.com/index.php?id=monkeys]http://lalalaala.com/index.php?id=monkeys[/url]
}else{
  echo 'that';
}

 

You may want to add isset as well so you don;'t get an error;

 

if(!isset($_GET['id']) && empty($_GET['id']))

[fortnox007]

 

You may want to add isset as well so you don;'t get an error;

 

if(!isset($_GET['id']) && empty($_GET['id']))

smart one