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

Posted by MUZ140886 Mandava
MUZ140886 Mandava
4,675 Points

challenge task 4 of 4

change the value in the flavor variable to cookie dough.preview the code and make the message appears

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

?>

2 Answers

Sjors Theuns
Sjors Theuns
6,091 Points
Sjors Theuns
Sjors Theuns
6,091 Points

Hi Madava,

I'm not sure what it is exactly that you want to know, but for the challenge you need something like:

<?php
$flavor = "strawberry";
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>";

?>

Above code is for challenge 3 of 4. When you change

$flavor = "strawberry";

into

$flavor = "cookie dough";

You can pass challenge 4 of 4.

Hopefully this makes it a bit clearer, but if you have any question let me know. For future questions please provide a bit more info.

MUZ140886 Mandava
MUZ140886 Mandava
4,675 Points

hi sjors

i did as you told me but it is reporting that: " it seems that challenge 2 is no longer visible"and i am struck on challenge since yesterday

Sjors Theuns
Sjors Theuns
6,091 Points
Sjors Theuns
Sjors Theuns
6,091 Points

Hi Madava,

I can redo this challenge over and over, without getting the message you are referring to. Perhaps it's better to contact support for this issue. It doesn't look like a programming error on which I can help.

Iain Diamond
Iain Diamond
29,379 Points
Iain Diamond
Iain Diamond
29,379 Points 
<?php
$flavor="Cheese";
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>";
}
?>

Change the value in the flavor variable to cookie dough

The challenge is to change $flavor so that the if condition becomes true, like this:

<?php
$flavor="cookie dough";
...

Hope this help :) iain