Isset

[Anon]

<?php
if(!isset($uname)) echo $uname;
}else{
echo ("I dunno your username");

?>

Sodoes anybody know what i did wrong here?
Thanks

[Anon]

Looks like you're saying 'if there is no uname display uname.
Try this instead

<?php
if(!isset($uname)) {
echo"I dunno your username";
} else {
echo $uname;
}

?>

Nick

[brandonschnell]

heres what you've got:
if($uname is not set) echo $uname;
else
echo "i dunno your username";

remove the NOT in your IF conditional
<?php
if(isset($uname)) echo $uname;
else echo "I dunno your username";
?>

[Anon]

Thanks Its working now!

[Anon]

Well, beside the fact i feel really dumb, for some reason i can't get this parse error out of here.. i tried a number of diffrent things but non of them worked.

<?php
if(isset($uname)) echo ("Welcome") $uname;
else echo "I dunno your username";
?> that is my code.
Thanks
Anthony

[Anon]

Hi
thry this, your $uname is out side of your echo statement, the '.' concatenates the var to the end of your string "contained in these"

<?php
if(isset($uname)) echo ("Welcome".$uname);
else echo "I dunno your username";
?>

[Anon]

Thanks works great! I didn't know that..
thanks again for your help
Anthony

[polson]

you may want to check out this tutorial on strings :

http://www.zend.com/zend/tut/using-strings.php

also, just a fyi, you can use ternery operator like so :

echo ($foo == 'bar') ? 'yes it is' : 'nope';

which is the same as :

if ($foo == 'bar') {
echo 'yes it is';
} else {
echo 'nope';
}

and regarding control-structure syntax such as if/else statements, check out :

http://php.net/manual/en/control-structures.php