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PHP PHP Basics Daily Exercise Program Conditionals

David Dunlop
David Dunlop
2,845 Points
Posted by David Dunlop
David Dunlop
2,845 Points

syntax error with if else conditional when checking studentOneGPA

this is the question I cant get right

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

/// <?php $studentOneName = 'Dave'; $studentOneGPA = 3.8;

$studentTwoName = 'Treasure'; $studentTwoGPA = 4.0;

//Place your code below this comment $NAME = $studentOneName; if ($studentOneGPA = 4.0) { echo "$NAME made the honor roll."; } else { echo "$NAME has a GPA of $STUDENT_GPA."; { $STUDENT_GPA = $studentTwoGPA; $NAME = $studentTwoName; if ($studentTwoGPA = 4.0) { echo "$NAME made the honor roll."; else { echo "$NAME has a GPA of $STUDENT_GPA."; } ?> ///

index.php 
<?php
$studentOneName = 'Dave';
$studentOneGPA = 3.8;

$studentTwoName = 'Treasure';
$studentTwoGPA = 4.0;

//Place your code below this comment

?>

1 Answer

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,972 Points
KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,972 Points

Some hints:

  • Check your brackets, particularly near your else statements
  • You don't need to create variables $NAME and $STUDENT_GPA. In fact I think the challenge fails if you don't just use the variables provided.
  • Check your comparison operators
  • Check case (ex: honor roll). In this challenge the text must be exact.