This is really weird. I have this piece of code.
$rewardAmt = $amt;
if(is_float($rewardAmt)){
print_r("is float");die;
} else {
print_r("is not float"); die;
}
value of $amt is 0.01. But it is going into else condition. So I did a var_dump of $amt. it says string(4) So I decided to typecast $amt
$rewardAmt = (float)$amt;
But the problem with this is even if the value of $amt is 1, it still gets typecast to float and goes into if condition, which shouldn't happen. Is there any other way to do this ? Thanks
Use filter_var()
with FILTER_VALIDATE_FLOAT
if (filter_var($amount, FILTER_VALIDATE_FLOAT))
{
// good
}
If you change the first line to
$rewardAmt = $amt+0;
$rewardAmt should be cast to a number.
You can use the unary +
operator, which will cast the string to the appropriate type ( int
or float
), and then test the resulting data type with is_float
:
$s = "3.00";
$n = +$s;
var_dump( $n ); // float(3)
var_dump( is_float($n) ); // true
$s = "3";
$n = +$s;
var_dump( $n ); // int(3)
var_dump( is_float($n) ); // false
You can check this by
$float = floatval($num); //Convert the string to a float
if($float && intval($float) != $float) // Check if the converted int is same as the float value...
{
// $num is a float
}else{
// $num is an integer
}
As suggested in the documentation of is_float function :
Note: To test if a variable is a number or a numeric string (such as form input, which is always a string), you must use is_numeric() .
The answer is using is_numeric() and not is_float(). This works also if the number tested is "0" or 0.
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