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tejaswini vadnal
1,587 Pointshis block of code displays two sentences related to flavors of ice cream. In this code challenge, we'll modify the code
<?php
echo "<p>Your favorite flavor of ice cream is "; echo "vanilla"; echo ".</p>"; echo "<p>Hal's favorite flavor is cookie dough, also!</p>"; $flavor ="tomato"; echo"your favorite flavor of ice cream is"; echo $flavor; echo "."; if($flavor == ".") echo "So my favorite flavor is, tomato"; echo $flavor; echo "tomato"
?>
<?php
echo "<p>Your favorite flavor of ice cream is ";
echo "vanilla";
echo ".</p>";
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
?>
2 Answers
jcorum
71,830 PointsYou need something along these lines:
$flavor = "oreo";
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
if ($flavor == "cookie dough") {
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
}
First, create the $flavor variable and assign it a value. Then echo it instead of the string vanilla. Then add an if statement that checks to see if $flavor is equal to "cookie dough".
jcorum
71,830 Pointstejaswini, not sure what the issue is for you, as this code is accepted by the editor:
<?php
$flavor = 'oreo';
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
if ($flavor == 'cookie dough') {
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
}
?>
tejaswini vadnal
1,587 Pointstejaswini vadnal
1,587 PointsThe message in the final echo command only makes sense if your favorite flavor is the same as Hal's. Add a conditional around that final echo command that checks if the flavor variable has a value of "cookie dough." (Remember to choose carefully between using a single equal sign and a double equal sign in the check.) Preview the code and, if you have a different flavor, make sure the message disappears.
how to solve this