With PHP, how to check if decimal number is above or less than 50?

Question

I am trying to check if a number which is inside a variable got a decimal number above 50 , or less than 50. Then depending on if it's 50 or higher, round the decimal number to 99 , and if it's less, round it to 00.

Here is a piece of my code:

public function roundPrice($price)
{
return round( (ceil($price) - 0.01), 2);
}

It makes all decimal numbers round up to 99 . I would need only decimals which are 50 or higher to be rounded in 99, and 49 or less become 00.

How could I achieve this in PHP ? Many thanks, I'm stuck here and don't know much how to do that.

do you mean 0.99 or actually 99? – Matthew Riches May 30 '12 at 14:34 — Matthew Riches, May 30 '12 at 14:34

score 5 · Accepted Answer · answered May 30 '12 at 14:41

On the off chance OP actually meant decimal places where 1.36 becomes 1.00 and 1.60 becomes 1.99.

Probably more elegant solutions, but here's one:

function roundPrice($price)
{
    $intVal = intval($price);
    if ($price - $intVal < .50) return (float)$intVal;
    return $intVal + 0.99;
}

http://codepad.org/5hIdQh4w

Sarfraz · Answer 2 · 2012-05-30T14:39:07.487

0

I would need only decimals which are 50 or higher to be rounded in 99, and 49 or less become 00.

Strange requirement but:

public function roundPrice($price)
{
  return $price >= 50 ? 99 : sprintf("%02s", 0);
}

For numbers greater than or equal to 50, it would return 99 and for numbers less than 50, it would return 00

edited May 30 '12 at 14:39

answered May 30 '12 at 14:33

Sarfraz

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I'm thinking he means decimal places. Ex 1.37 becomes 1.00 and 1.67 becomes 1.99? – tigrang May 30 '12 at 14:34
1

I think you missed a ? there... :) – Alex Howansky May 30 '12 at 14:34
Thanks guys. @tigrang: that is correct, sorry my question wasn't clear enough now that I read it again. – mlclm Jun 01 '12 at 14:47

score 0 · Answer 3 · answered May 30 '12 at 14:34

0

You're not really rounding, so just be direct about it.

public function roundPrice($price)
{
    if($price >= 50)
    {
        return 99;
    }
    else
    {
        return 0;
    }
}

Your stated problem does not handle the $price being equal to 50. I made the assumption that the value of exactly 50 rounds up to 99.

answered May 30 '12 at 14:34

Surreal Dreams

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`return 0;` would return just `0` not `00` – Sarfraz May 30 '12 at 14:34
@Sarfraz - 0 is a number, not a string. Do you want a string? – ghoti May 30 '12 at 14:35
0 <> 00? Considering the statement of the problem, I'm assuming that 0 = 00. – Surreal Dreams May 30 '12 at 14:35
@SurrealDreams: True but it seems OP wants two zeroes although they are same. – Sarfraz May 30 '12 at 14:36
I can make it say `return 00`, but that won't make it behave any differently than it does. – Surreal Dreams May 30 '12 at 14:37

score 0 · Answer 4 · answered May 30 '12 at 14:35

0

This modifies your function to return the decimal value if it's 50 or above and 0 otherwise.

public function roundPrice($price)
{
   $decValue = round( (ceil($price) - 0.01), 2);

   if ($decValue >= 50) return $decValue;
   else return 0;
}

answered May 30 '12 at 14:35

nageeb

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Kjir · Answer 5 · 2012-05-30T14:49:44.747

0

I think you want this:

public function roundPrice($price) {
    $p = round($price, 0);
    if( ceil($price) == $p) {
        return $p - 0.01;
    }
    return $p;
}

This returns X.99 for values >= X.5 and X.00 for all other values.

edited May 30 '12 at 14:49

answered May 30 '12 at 14:43

Kjir

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With PHP, how to check if decimal number is above or less than 50?

5 Answers5