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PHP Build a Simple PHP Application Creating the Menu and Footer Variables and Conditionals

Mohamed Hafez
Mohamed Hafez
1,668 Points
Posted by Mohamed Hafez
Mohamed Hafez
1,668 Points

I am not sure where I went wrong here help please

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

Mohamed Hafez
Mohamed Hafez
1,668 Points
Mohamed Hafez
Mohamed Hafez
1,668 Points

it asks me to change the variable value from vanilla to cookie dough which I've completed however says I am failing is this a bug ? in my head ?

2 Answers

Prathom Satapronpinyo
Prathom Satapronpinyo
12,759 Points
Prathom Satapronpinyo
Prathom Satapronpinyo
12,759 Points

Here is the correct answer

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

?>

I think, it nothing to do with php syntax but it should be how the program check your answer. u have to get rid of a dot after the word "cookie dough" and add paragraph syntax back to your code.

hope this might help.

Jovanny Elias
Jovanny Elias
16,204 Points
Jovanny Elias
Jovanny Elias
16,204 Points

Make sure you have 

echo ".</p>";

instead of

 echo ".";

php is very strict when it comes to having the correct syntax unlike other languages that still work even when forgetting to add certain characters or misspelled words. Give this a try it should work