Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 2 down vote favorite
1

I've been working on this bit of code for about an hour and can't seem to figure out why it's not working. Does PHP allow If/Else statements within While loops? I've tried adding different echos to both If/Else statements and the latter (Else) will not show. Is this because I'm trying to use the same variable names? 

while($row = mysql_fetch_array($result))

{

//assign variables
$title = $row['title'];
$file_url = $row['file_location'];
$category = $row['category'];
$layout = $row['layout'];


    If ($layout = "vertical")       
    {
        //Page Layout
        $BODYLAYOUT = "vertical_body";
        $GAMECONTAIN = "vertical_gameContain";
        $GAMEWIDTH = "vertical_game";
    }
    Else
    {
        // Page Layout
        $BODYLAYOUT = "horizontal_body";
        $GAMECONTAIN = "horizontal_gameContain";
        $GAMEWIDTH = "horizontal_game";
    }
share|improve this question
there is a spelling mistake in else key word , It should be 'else' , not 'Else' – aravind.udayashankara Dec 9 '12 at 2:09
PHP is case-insensitive in regards to that. – cryptic ツ Dec 9 '12 at 2:09
Yeah , I agree , with you , but why am I getting a fatel error if I execute this on that line number as PHP depricated : some thing like that – aravind.udayashankara Dec 9 '12 at 2:14
and also you answer here is correct , but still I felt , even this might be a reason – aravind.udayashankara Dec 9 '12 at 2:14
I'm not getting such an error. Can you post the exact output of the error? – cryptic ツ Dec 9 '12 at 2:19
show 1 more comment

1 Answer

up vote 3 down vote accepted 
if ($layout = "vertical")

should be:

if ($layout == "vertical")

Otherwise you are assinging a value to $layout of 'vertical' opposed to comparing it's value to see if it's equal to 'vertical'. That assignment will otherwise equal to true, reason the first part runs and the ELSE does not.

One method I use to prevent accidents like this is to put the constant first such as:

if ("vertical" == $layout)

That way if I miss the other = sign PHP will error, rather than assign the value erroneously.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.