Php Variable Problems

[paulmason411 0]

Hi guys,

I'm new to php and I think i might be having problems with my syntax.

Heres a snippet of code that doesn't seem to work:

<?php $csslayout = 1; ?>					   <div id="forumposts">  <?php mosLoadModules ( 'user1' ); ?> /* can have the code <?php $csslayout = 2; ?> */</div><?php if (csslayout == 1){ ?>  <div id="content">	<?php mosMainBody(); ?>  </div><?php } ?><?php if (csslayout == 2){ ?>  <div id="content2">	<?php mosMainBody(); ?>  </div><?php } ?>

It never gets past the if statement.

What am I doing wrong?

[BuffaloHelp 24]

You were missing $ when calling your variable.

Try this:

<?php $csslayout = "1";					   echo "<div id='forumposts'>";mosLoadModules ("user1");echo "</div>";if ($csslayout == "1") { 	  echo "<div id='content'>";	  mosMainBody();	  echo "</div>";	}if ($csslayout == "2") { 	  echo "<div id='content2'>";	  mosMainBody();	  echo "</div>";	}?>

You can see that I've switched double quote with single quote when using PHP command that uses double quote. There is another way to use single quote command so that you can use double quote inside the PHP command. But try this first to see if it's working.

A neat trick I learned the first is the use of ECHO and PRINT command. By using this, you don't have to switch in and out from HTML and PHP.

[fffanatics 0]

Well the code above should work. The reason yours was not working was that you were <?php and ?> after every php command. Your if statement was being opening and then you would stop using php and go to html. This would be seen in the server as an if statement without a body and would leave you with a brace not matched. If you are working with conditionals or loops in php make sure that you have the whole conditional or loop in the same php block or else you risk it not working correctly. To ensure you do this use the echo and print commands (like stated above) to send html to the page from the php block.

[QuickSilva 0]

The reason yours was not working was that you were <?php and ?> after every php command.

Not true, the reason he was doing this is to escape the PHP so he could put the HTML in. This did not cause the error.

Example:

<?php$var = "hi";if ($var == "hi"){ //if the variable $var is equal to "hi" carry on.echo 'Hey there!';}?>

Is the same as:

<?php$var = "hi";if ($var == "hi"){ //if the variable $var is equal to "hi" carry on.?>Hey there!<?php}?>

Also I think this is the other way to echo:

<?phpecho <<<EOFyou can put anything here really including " and 'EOF;?>

Hope this helps!

 

-Tom