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Start your free trialCoco Works
3,294 PointsConditionals in PHP
Hello!
I am stuck on this one question, and I can't seem to figure it out.
In the exercise for Conditionals in the PHP course, the instructions are as follows:
"Check if each student has a GPA of 4.0.
If the student has a GPA of 4.0, use the student name variable to replace NAME when displaying the following line:
NAME made the Honor Roll
ELSE If the students GPA is not equal to 4.0, within an else block, use the student name variable to replace NAME AND the student GPA variable to replace STUDENT GPA when displaying the following line:
NAME has a GPA of STUDENT GPA"
When typing out my code, the preview prints out the results in one line. When clicking on "Check Work" I get the message: "Bummer: Use the variables to display the message"
I am unsure which variables it is referring to, as I was confident to have used the required ones. But, clearly, I am doing something wrong. Can anyone point me in the right direction?
<?php
$studentOneName = 'Dave';
$studentOneGPA = 3.8;
$studentTwoName = 'Treasure';
$studentTwoGPA = 4.0;
//Place your code below this comment
if ($studentOneGPA == 4.0){
echo "$studentOneName made the Honor Roll\n";
}
else{
echo "$studentOneName has a GPA of $studentOneGPA\n";
}
if ($studentTwoGPA == 4.0){
echo "$studentTwoName made the Honor Roll\n";
}
else{
echo "$studentOneName has a GPA of $studentOneGPA\n";
}
?>
2 Answers
Joseph Yhu
PHP Development Techdegree Graduate 48,638 PointsLook at your last else
block, $studentOneName
and $studentOneGPA
should be $studentTwoName
and $studentTwoGPA
.
Coco Works
3,294 PointsOh my! That is so embarrassing. Lesson learned: don't be lazy and copy and paste mindlessly. Thanks for the quick answer, appreciate that.