I use the following code in php to check if the number is integer:
$qtd_bo = "1.20";
if (!preg_match('/^[1-9][0-9]*$/', $qtd_bo)) {
echo "Não é inteiro";
}
It returns me that number 1.20 is not integer. Until then, okay.
The problem is that if I put 1.00 it tells me that it is not integer.
I need to check if there is any value after the "."
The value 1.00 will always be considered a float , however you can validate it as follows:
$value = "1.00";
return floor($value) != $value;
If the value in float actually is an "integer", the condition will return TRUE , because PHP compares the values without taking into account the type and function floor returns just the whole part, just.
Number 1 is int but 1.0 is not int , it is float or it can also be double or decimal depending on the language. Did you understand the difference?
So 1.0 or 1.0000, will never be a int .
Test with a test
$str = 1.00;
var_dump(gettype($str)); // retorna double
var_dump(is_int($str)); // retorna false
An addendum, beware of this regular expression, since -1 is an integer, but the expression returns false:
if (!preg_match('/^[1-9][0-9]*$/', $str)) {
echo 'não é inteiro';
}
Testing negative integer
$str = -1;
var_dump(gettype($str)); // retorna integer
var_dump(is_int($str)); // retorna true
If you want to do with regex :
<?php
$testes = array('-65','065','16as321','132,16','16544.01','-1');
foreach ( $testes as $valor ){
printf( "%s %s um número inteiro\n" , $valor , preg_match( '/^\-?[1-9][0-9]*$/' , $valor ) ? 'é' : 'não é' );
}
?>
See working at Ideone .
In PHP , there is is_int , which informs if the variable is of the integer type.
<?php
if (is_int(23)) {
echo "is integer\n";
} else {
echo "is not an integer\n";
}
var_dump(is_int(23));
var_dump(is_int("23"));
var_dump(is_int(23.5));
var_dump(is_int(true));
?>