PHP：检查输入是否为有效数字的最佳方法？
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PHP: Best way to check if input is a valid number?
提问by medusa1414
What is the best way of checking if input is numeric?
检查输入是否为数字的最佳方法是什么？
- 1-
- +111+
- 5xf
- 0xf
- 1-
- +111+
- 5xf
- 0xf
Those kind of numbers should not be valid. Only numbers like: 123, 012 (12), positive numbers should be valid. This is mye current code:
那些数字不应该是有效的。只有像：123, 012 (12) 这样的数字，正数才有效。这是我当前的代码：
$num = (int) $val;
if (
preg_match('/^\d+$/', $num)
&&
strval(intval($num)) == strval($num)
)
{
return true;
}
else
{
return false;
}
回答by
ctype_digitwas built precisely for this purpose.
ctype_digit正是为此目的而建造的。
回答by Mathieu Dumoulin
I use
我用
if(is_numeric($value) && $value > 0 && $value == round($value, 0)){
to validate if a value is numeric, positive and integral
验证值是否为数字、正数和整数
I don't really like ctype_digit as its not as readable as "is_numeric" and actually has less flaws when you really want to validate that a value is numeric.
我真的不喜欢 ctype_digit，因为它不像“is_numeric”那样可读，而且当你真的想验证一个值是数字时，它实际上有更少的缺陷。
回答by John Conde
$options = array(
'options' => array('min_range' => 0)
);
if (filter_var($int, FILTER_VALIDATE_INT, $options) !== FALSE) {
// you're good
}
回答by rdlowrey
return ctype_digit($num) && (int) $num > 0
回答by MD. Shafayatul Haque
For PHP version 4 or later versions:
对于 PHP 4 或更高版本：
<?PHP
$input = 4;
if(is_numeric($input)){ // return **TRUE** if it is numeric
echo "The input is numeric";
}else{
echo "The input is not numeric";
}
?>
回答by Kareem
The most secure way
最安全的方式
if(preg_replace('/^(\-){0,1}[0-9]+(\.[0-9]+){0,1}/', '', $value) == ""){
//if all made of numbers "-" or ".", then yes is number;
}
