CODEWITHSUNDEEP
phpif-statementrange
three3
three3

Reputation: 2846

Checking the Range of a Number in PHP

I have a simple PHP script that is checking the range of a number. For some reason, once the number I am checking equals 100% the code does not work. Here is my code:

$percent_completed = '100%';                            

if ($percent_completed <= '20%') {
    $color = 'danger';
} elseif ($percent_completed >= '21%' && $percent_completed < '40%') {
    $color = 'warning';
} elseif ($percent_completed >= '40%' && $percent_completed < '60%') {
    $color = 'info';
} elseif ($percent_completed >= '60%' && $percent_completed < '80%') {
    $color = 'primary';
} elseif ($percent_completed >= '80%' && $percent_completed < '100%') {
    $color = 'default';
} else {
    $color = 'success';
}

echo $color;

All of the conditional checks above work just fine until $percent_completed is equal to 100%. For some reason it is set to 100% the $color that is printed out is danger. I am sure it is a simple fix but everything I have tried does not work.

Upvotes: 1

Views: 1470

Answers (5)

Aycan Yaşıt
Aycan Yaşıt

Reputation: 2104

Just to point out another approach, I'm posting an alternative solution. Sometimes it's more readable to use a simple algebric formula instead of plenty of if-else conditions.

//Assign current percent value to a variable
$percent_completed = 100;
//Assign an array of all notifications
$array_notifications = array("danger", "warning", "info", "primary", "default", "success");
//Calculate index of current notification
$current_index = floor($percent_completed / 20);
//Print or do something else with detected notification type
echo $array_notifications[$current_index];

Upvotes: 2

Halcyon
Halcyon

Reputation: 57709

Comparing percentage strings could work if you're using a natural-order compare function.

Natural order compare functions separate strings and numbers and will treat the numbers as numbers rather than string.

So instead of:

[ "1", "10", "2" ]

You will get:

[ "1", "2", "10" ]

Some languages have this function built in (like PHP: strnatcmp) but JavaScript sadly does not. Writing your own implementation is not very hard, but it's not very easy either.

In this case I would definitely recommend simplifying (like John's solution).

Upvotes: 1

brandonscript
brandonscript

Reputation: 72855

You could simplify this considerably.

  • Remove the quotes (numbers are integers)
  • Remove the % signs from your percentages for calculation
  • php processes the first successful if statement correctly and exits the condition. Stack from top to bottom, and you use 1/2 the code. (Or you could write it bottom to top and use > instead).
$p = 100;                            
if ($p == 100)
    $color = "success";
elseif ($p >= 80)
    $color = "default";
elseif ($p >= 60)
    $color = "primary";
elseif ($p >= 40)
    $color = "info";
elseif ($p > 20)
    $color = "warning";
elseif ($p <= 20)
    $color = "danger";
echo $color;

Upvotes: 1

John Conde
John Conde

Reputation: 219804

Get rid of the % from your $percent_completed variable. It makes it as string which will give you different results than when comparing as a integer (for numbers). 

$percent_completed = 100;                            

if ($percent_completed <= 20) {
    $color = 'danger';
} elseif ($percent_completed < 40) {
    $color = 'warning';
} elseif ($percent_completed < 60) {
    $color = 'info';
} elseif ($percent_completed < 80) {
    $color = 'primary';
} elseif ($percent_completed < 100) {
    $color = 'default';
} else {
    $color = 'success';
}

echo $color;

Upvotes: 8

Ryan
Ryan

Reputation: 3582

You're doing calculations on a string.

That means that '2%' is actually higher than '100%' (for example).

Remove the percentage symbol and use it when required during output.

Upvotes: 4

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