Build a simple PHP application > Creating the menu and footer > Code Challenge: Variables and Conditionals > Task: 3/4

Here's a hint, the question is asking you to echo the variable flavor not the literal text.

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5 to 10 words is generally a good length.

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James thanks for the note. btw i can't pass the task 3 #Help

@Mohamed -

Post the code you have so far into a PHP playground.

Done

I posted my code there but realy i don't get the next step, after that , what i'm gonna do?

What's the link to the PHP code you posted?

http://codepad.viper-7.com/6oLFtk

Now you need a conditional to test to see if the variable flavor is equal to cookie dough and if so output Randy's favorite flavor is cookie dough, also..

Otherwise if the flavor is equal to something else, writing "also" at the end of the sentence would not be grammatically correct, so the string should instead say Randy's favorite flavor is cookie dough.

Let us know how you are getting along and if you need any more help.

@James -

i have done , here is the link

http://codepad.viper-7.com/OcpQlc

thanks for the helping :)

Hey @Mohamed,

That code is close but not quite correct. Line 4 should look like this:

echo $flavor;

In line 2, you set the variable to have a value. In line 4, you are supposed to echo out whatever value is in the variable. Does that make sense?

You need an else because if I change the value assigned to flavor to something other than cookir dough it now no longer print out what Randy's favorite flavor is.

Actually, James, that's the desired behavior. The code challenge says to display the message about my favorite flavor if our favorite flavors are the same. This paragraph wouldn't make any sense:

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

The conditional should make sure the second sentence only shows when they are the same.

@Randy - Oops, that's what I get for not going back and checking the the instructions on the code challenge.