Conditional Code

You've got too many PHP opening and closing tags. You don't want to be echoing out the if statement because you can't, instead you want to echo the result of the if statement which you already are.

I forgot that it's initially taught using echo statements with no concatenation. This is how the first, correct, part of that would look concatenated which is much nicer to read:

<?php 
$flavor = "chocolate"; 
echo "Your favorite flavor of ice cream is: " . $flavor . "."; 
?>

It's the bit after that that gets confused and messy. You've just got too many of some things in there.

The only code in the entire statement i changed was to replace this sentence below which was enclosed with p tags: Randy's favorite flavor is cookie dough, also!

with this which I placed with in the original p tags:

<?php if ($flavor == "cookie dough") { echo "Randy's favorite flavor is cookie dough also!"; } ?>

I was not attempting to echo the information contained within the variable. The challenge was to make a conditional statement that would result in not showing, "Randy's favorite flavor is cookie dough, also!" sentence if my favorite flavor ice cream was not cookie dough.

So my question is not how to echo the variable, but what is wrong with my if statement? Obviously my $flavor is chocolate, so the statement should not show up. My output is creating a parse error. Any idea why?

Also, the code didn't look like that until I pasted it into the comment. It was very neat and easy to read...

Taking out the bit that's right and works, tell me what's wrong with the following:

<?php 

echo "<?php if ($flavor == "cookie dough") { echo   "Randy's favorite flavor is cookie dough also!"; } ?>";

?>

You're asking me the same question I asked. If I knew what was wrong, I wouldn't have posted my question.

I'm trying to get you to work out what's wrong with it rather than just give the answer. There's no point me just telling you because you won't have learnt anything.

Count how many <?php and ?> there are in that statement and read what I wrote about echoing out the if statement and echoing out the result.

You don't need to echo the if FUNCTION, just the result. echo and if are both parts of PHP and are already in a PHP block because you've got opening <?php before the variable declaration.

Making anymore sense?

I changed it to the following and it worked:

if ($flavor == "cookie dough") { echo "Randy's favorite flavor is cookie dough also!"; }

Of course the next challenge tells you to change the variable to cookie dough and see what happens. When I did, it says it looks like task three is no longer passing. Very frustrating...time to forget this one and move on. I can't see how changing the variable from chocolate to cookie dough would cause it not to "pass". It should just show the message.

I forgot to mention, I tried to work it out for about an hour before I finally asked. Not one to ever ask for help, I only asked after I was ready to give up. I didn't need to be made to feel like something of an idiot for doing so.

You shouldn't feel like you were made to feel like an idiot. In the forum, we prefer to put things in different perspectives and offer hints, tips and coax towards the right answer rather than just give it out. If the correct answer was just given out all the time, there's no point in having code challenges as a quick forum search would yield the answer. Nobody learns that way.

So, you've worked out what was wrong and adjusted it which is excellent and the whole point of this learning environment. :)

And you're correct, copying and pasting cookie dough from the if statement to be the value of $flavor should mean the statement resolves to true and thus show the result which is the echoed out string.

I am still working on this....Although I did the if statement correctly and all was fine for part 3 of 4, now the last part of the code challenge says to change the variable to match cookie dough, which I did. I reread the block of code and everything looks okay with the only difference being my $chocolate variable now reads $cookie dough. For some reason I keep getting a message that it is no longer parsing. I only changed $chocolate to $cookie dough and touched nothing else. Why I am getting an error?

Here is the code:

<?php
$flavor = "cookie dough";
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
if ($flavor == "cookie dough") { 
  echo "Randy's favorite flavor is cookie dough, also!"; 
}

?>

Finally! Thank you Jason!

Do I have to echo the p tags? I did, but I wasn't sure.

<?php
$flavor = "cookie dough";
echo "<p>Your favorite flavor of ice cream is ";
echo "$flavor";
echo ".</p>";
echo "<p>";
if ($flavor == "cookie dough") { 
echo "Randy's favorite flavor is cookie dough, also!";
}
echo "</p>";

?>

Thank you Jason. I didn't think it looked right with the p tags echoed. It did pass the code challenge, but I will remember to do it differently in the future.

Thanks again for the response.

Kelly