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PHP PHP Basics Daily Exercise Program Conditionals

Alex Bauer
Alex Bauer
9,426 Points
Posted by Alex Bauer
Alex Bauer
9,426 Points

How do I display the value of the variable $studentOneGPA correctly?

I thought that double quotes would enable me to use the value of the variable instead of the name of the variable but it gives me a syntax error. I'm supposed to display the name of the each student and their GPA if it's lower than 4.0.

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

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

//Place your code below this comment
if $studentOneGPA == 4.0 {
  echo $studentOneName . '' . 'has made the Honor Roll'
} else {
  echo $studentOneName . '' . "has a GPA of $studentOneGPA"
if $studentTwoName == 4.0 {
  echo $studentTwoName . '' . 'has made the Honor Roll'
} else {
  echo $studentTwoName . '' . "has a GPA of $studentTwoGPA"
?>

1 Answer

Sir Dev
Sir Dev
247 Points

You forget to wrap your if statements with parentheses, the condition that your are testing must be encloses in parentheses and you also left some curly braces out as well. I took the liberty of fixing your code its below:

$studentOneName = 'Dave';
$studentOneGPA = 3.8;

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

if ($studentOneGPA == 4.0) {
  echo $studentOneName . '' . 'has made the Honor Roll';
} else {
  echo $studentOneName . '' . "has a GPA of $studentOneGPA";
}
if ($studentTwoName == 4.0) {
  echo $studentTwoName . '' . 'has made the Honor Roll';
} else {
  echo $studentTwoName . '' . "has a GPA of $studentTwoGPA";
}
Alex Bauer
Alex Bauer
9,426 Points
Alex Bauer
Alex Bauer
9,426 Points

Okay thank you so much for your help! I really appreciate it!