[英]PHP check if is integer
I have the following calculation: 我有以下计算:
$this->count = float(44.28)
$multiple = float(0.36)
$calc = $this->count / $multiple;
$calc = 44.28 / 0.36 = 123
Now I want to check if my variable $calc
is integer (has decimals) or not. 现在我想检查我的变量$calc
是否为整数(有小数)。
I tried doing if(is_int()) {}
but that doesn't work because $calc = (float)123
. 我尝试了if(is_int()) {}
但是这不起作用,因为$calc = (float)123
。
Also tried this- 也试过这个 -
if($calc == round($calc))
{
die('is integer');
}
else
{
die('is float);
}
but that also doesn't work because it returns in every case 'is float'
. 但这也行不通,因为它在每种情况下都返回'is float'
。 In the case above that should'n be true because 123 is the same as 123 after rounding. 在上面的情况下,这应该是真的,因为在舍入后123与123相同。
As CodeBird pointed out in a comment to the question, floating points can exhibit unexpected behaviour due to precision "errors". 正如CodeBird在对该问题的评论中指出的那样,由于精度“错误”,浮点可能会出现意外行为。
eg 例如
<?php
$x = 1.4-0.5;
$z = 0.9;
echo $x, ' ', $z, ' ', $x==$z ? 'yes':'no';
prints on my machine (win8, x64 but 32bit build of php) 在我的机器上打印(win8,x64但是32位构建的php)
0.9 0.9 no
took a while to find a (hopefully correct) example that is a) relevant to this question and b) obvious (I think x / y * y
is obvious enough). 花了一段时间才找到一个(希望是正确的)例子,这个例子是a)与这个问题相关,b)显而易见(我认为x / y * y
很明显)。
again this was tested on a 32bit build on a 64bit windows 8 再次,这是在64位Windows 8上的32位构建上测试的
<?php
$y = 0.01; // some mambojambo here...
for($i=1; $i<31; $i++) { // ... because ...
$y += 0.01; // ... just writing ...
} // ... $y = 0.31000 didn't work
$x = 5.0 / $y;
$x *= $y;
echo 'x=', $x, "\r\n";
var_dump((int)$x==$x);
and the output is 而输出是
x=5
bool(false)
Depending on what you're trying to achieve it might be necessary to check if the value is within a certain range of an integer (or it might be just a marginalia on the other side of the spectrum ;-) ), eg 根据您要实现的目标,可能需要检查该值是否在整数的某个范围内(或者它可能只是频谱另一侧的边际;-)),例如
function is_intval($x, $epsilon = 0.00001) {
$x = abs($x - round($x));
return $x < $epsilon;
};
and you might also take a look at some arbitrary precision library, eg the bcmath extension where you can set "the scale of precision". 您还可以查看一些任意精度库,例如bcmath扩展 ,您可以在其中设置“精度等级”。
round()
will return a float. round()
将返回一个浮点数。 This is because you can set the number of decimals. 这是因为您可以设置小数位数。
You could use a regex: 你可以使用正则表达式:
if(preg_match('~^[0-9]+$~', $calc))
PHP will convert $calc
automatically into a string when passing it to preg_match()
. 当将$calc
传递给preg_match()
时,PHP会自动将$calc
转换为字符串。
You can do it using ((int) $var == $var)
你可以用((int) $var == $var)
来做
$var = 9;
echo ((int) $var == $var) ? 'true' : 'false';
//Will print true;
$var = 9.6;
echo ((int) $var == $var) ? 'true' : 'false';
//Will print false;
Basically you check if the int value of $var
equal to $var
基本上你检查$var
的int值是否等于$var
You can use number_format()
to convert number into correct format and then work like this 您可以使用number_format()
将数字转换为正确的格式,然后像这样工作
$count = (float)(44.28);
$multiple = (float)(0.36);
$calc = $count / $multiple;
//$calc = 44.28 / 0.36 = 123
$calc = number_format($calc, 2, '.', '');
if(($calc) == round($calc))
die("is integer");
else
die("is not integer");
Ok I guess I'am pretty late to the party but this is a alternative using fmod() which is a modulo operation. 好吧,我想我已经很晚了,但这是一个使用fmod()的替代方案,这是一个模运算。 I simply store the fraction after the calculation of 2 variables and check if they are > 0 which would imply it is a float. 我只是在计算2个变量后存储该分数,并检查它们是否> 0,这意味着它是一个浮点数。
<?php
class booHoo{
public function __construct($numberUno, $numberDos) {
$this->numberUno= $numberUno;
$this->numberDos= $numberDos;
}
public function compare() {
$fraction = fmod($this->numberUno, $this->numberDos);
if($fraction > 0) {
echo 'is floating point';
} else {
echo 'is Integer';
}
}
}
$check= new booHoo(5, 0.26);
$check->compare();
Edit: Reminder Fmod will use a division to compare numbers the whole documentation can be found here 编辑:提醒Fmod将使用一个部门来比较数字, 这里可以找到整个文档
if (empty($calc - (int)$calc))
{
return true; // is int
}else{
return false; // is no int
}
Try this: 试试这个:
//$calc = 123;
$calc = 123.110;
if(ceil($calc) == $calc)
{
die("is integer");
}
else
{
die("is float");
}
you may use the is_int()
function at the place of round()
function. 你可以在is_int()
函数的地方使用is_int()
round()
函数。
if(is_int($calc)) {
die('is integer');
} else {
die('is float);
}
I think it would help you 我认为这会对你有所帮助
A more unorthodox way of checking if a float is also an integer: 检查浮点数是否也是整数的更为非正统的方法:
// ctype returns bool from a string and that is why use strval
$result = ctype_digit(strval($float));
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