Can anyone find the problem?

Hi Andrew,

It looks like you have two issues here, one is when you're setting the variable $flavor, the other is when you are trying to echo that same variable.

When you set a variable to be a string, you need to surround that string with double quotes. So it should look like this:

$flavor = "cookie dough";

When you are trying to access the value of a variable, like when you echo one, you do not need to surround it with quotes, if you do then php will think it is a string and not a variable. So it should look like this:

echo $flavor;

The rest of your code looks good. Hope this helps!

Hey Andrew,

Michael's right. Right now, the code isn't valid PHP. Putting "cookie dough" in quotation marks will fix that.

Once you get the code valid, you should encounter one other problem. I intended for the whole second sentence to be inside the conditional. Here are the two scenarios:

(1) If your favorite flavor is something like rocky road, the page should display this:

Your favorite flavor of ice cream is rocky road.

(2) If your favorite flavor is cookie dough, the page should display this:

Your favorite flavor of ice cream is cookie dough. Randy's favorite flavor is cookie dough, also! 

(I suspect you had that right in an earlier attempt; otherwise, you probably wouldn't have passed step three.)

@Michael + @Randy woooh teamwork!