PHP basic: why my variable showing wrong value? - php

Question

I am new to PHP. I am trying to create a simple form where user will submit there name and mark they got in the exam.
<form action="" method="post">
<input type="text" name="name" id="" placeholder="Your name">
<input type="number" name="mark" id="" placeholder="Your Mark">
<input type="submit" name="submit" value="Submit">
</form>
<?php
if(isset($_POST["submit"])){
$name = $_POST['name'] ;
$mark = $_POST['mark'];
if( $mark > 80 ){
$result = "A+";
}elseif($mark > 70 && $mark < 80){
$result = "A";
}
elseif($mark > 60 && $mark < 70){
$result = "A-";
}
elseif($mark > 40 && $mark < 60){
$result = "B";
}else{
$result = "Fail";
}
echo $result;
}
?>
But if I echo the result variable it is always showing 'Fail'. What is I am doing wrong? thanks

Answer 1

As mentioned in the comments, when comparing ranges, you need to consider the corner cases:
$mark >= 70 && $mark < 80
Also there are few ways to simplify your code: If you check for marsk >= 80 in the first IF, there is no need to check for that in the next elseif:
if( $mark >= 80 )
{
$result = "A+";
}
elseif($mark >= 70)
{
$result = "A";
}
elseif($mark >= 60)
{
$result = "A-";
}
elseif($mark >= 40)
{
$result = "B";
}
else
{
$result = "Fail";
}
echo $result;