php 使用PHP检查变量是否包含正整数的最短方法？

Shortest way to check if a variable contains positive integer using PHP?

提问by Naveed

I want to check if user input a positive integer number.

我想检查用户是否输入了一个正整数。

1    = true
+10  = true
.1   = false
-1   = false
10.5 = false


Just a positive number. 
No characters.
No special character.
No dot.
No minus sign.

I tried is_int()function but it is returning false even on positive integers. Is there a string to intproblem?

我尝试过is_int()函数，但即使在正整数上它也返回 false。是否有字符串到 int 的问题？

回答by Yacoby

Something like this should work. Cast the value to an integer and compare it with its original form (As we use ==rather than ===PHP ignores the type when checking equality). Then as we know it is an integer we test that it is > 0. (Depending on your definition of positive you may want >= 0)

像这样的事情应该有效。将值转换为整数并将其与其原始形式进行比较（因为我们使用==而不是===PHP 在检查相等性时忽略类型）。然后我们知道它是一个整数，我们测试它是否 > 0。（取决于您可能想要的正数定义>= 0）

$num = "20";

if ( (int)$num == $num && (int)$num > 0 )

回答by Gordon

Try the native Filter function*

尝试原生过滤器功能*

filter_var($value, FILTER_VALIDATE_INT, array(
    'options' => array('min_range' => 1)
));

^{* if you just want to make sure the input string consists of an arbitrary length digit sequence, use a regex with [0-9] or [\d+]}

^{* 如果您只想确保输入字符串由任意长度的数字序列组成，请使用带有 [0-9] 或 [\d+] 的正则表达式}

Examples with filter_var:

示例filter_var：

var_dump( filter_var(1, FILTER_VALIDATE_INT) ); // int(1)

var_dump( filter_var('1', FILTER_VALIDATE_INT) ); // int(1)

var_dump( filter_var('+10', FILTER_VALIDATE_INT) ); // int(10)

var_dump( filter_var(.1, FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var('.1', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var(-1, FILTER_VALIDATE_INT, 
    array('options' => array('min_range' => 1))) ); // bool(false)

var_dump( filter_var('-1', FILTER_VALIDATE_INT, 
    array('options' => array('min_range' => 1))) ); // bool(false)

var_dump( filter_var('2147483648', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var('0xFF', FILTER_VALIDATE_INT) ); // bool(false)

var_dump( filter_var(0xFF, FILTER_VALIDATE_INT) ); // int(255)

回答by Decko

I would say this is the best way

我会说这是最好的方法

if (is_int($num) && $num > 0)

as typecasting to an int is very slow.

因为类型转换为 int 非常慢。

回答by Imabigfatseacow

I use a regular expression. Very simple if you think about it. You need more punctuation marks if you want to make a number not a whole positive integer (minus sign and a period). So this just makes sure the only thing you have are numbers 0-9 for the value.

我使用正则表达式。如果你考虑一下，这很简单。如果你想让一个数字不是一个完整的正整数（减号和一个句点），你需要更多的标点符号。因此，这只是确保您拥有的唯一值是数字 0-9。

if(!ereg('^[0-9]+$', $value)) {
  $errors .= "This is not a positive whole number";
}

You could add another part on there to make sure it is less than a certain amount of characters as well. Hope this helps.

您可以在那里添加另一部分以确保它也少于一定数量的字符。希望这可以帮助。

回答by Nopcea Flavius

the easiest way is:

最简单的方法是：

if intval($x) > 0 echo "true"

如果 intval($x) > 0 回声“真”

回答by hram908

if(!preg_match('/^[0-9]+$/', $input)) {

if(!preg_match('/^[0-9]+$/', $input)) {

Deprecated: Function ereg() is deprecated

已弃用：不推荐使用函数 ereg()