I'm not sure if this is intended behaviour or not, but I can see that is might be. Just thought I would bounce this to see what people think about it:
<?php if (false) ?>Hello<?php else echo "Hi"; echo " World\n"; ?> This returns a parse error due to the fact there is no brackets around the first part of the if statement. It seems that because the next line after the if is not php, then it terminates the if at the closing of the php tag. But if you do this: <?php if (false) ?>Hello<?php ?> or <?php if (true) ?>Hello<?php ?> The word "Hello" is always output for the same reason. I can understand why to some extent in the way that is behaves, but it depends on your point of view as to what "statement" (as per the manual) actually means. Why can the bracketed version of if conditionaly display html, yet the unbracketed lazy man version cant? Any ideas as to whether or not this is supposed to happen like this or not? Tim -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
