php 检查变量是否为整数
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php check to see if variable is integer
提问by Arian Faurtosh
Is there a nicer way to do this?
有没有更好的方法来做到这一点？
if( $_POST['id'] != (integer)$_POST['id'] )
echo 'not a integer';
I've tried
我试过了
if( !is_int($_POST['id']) )
But is_int()doesn't work for some reason.
但is_int()由于某种原因不起作用。
My form looks like this
我的表格看起来像这样
<form method="post">
<input type="text" name="id">
</form>
I've researched is_int(), and it seems that if
我研究过is_int()，似乎如果
is_int('23'); // would return false (not what I want)
is_int(23); // would return true
I've also tried is_numeric()
我也试过 is_numeric()
is_numeric('23'); // return true
is_numeric(23); // return true
is_numeric('23.3'); // also returns true (not what I want)
it seems that the only way to do this is:[this is a bad way, do not do it, see note below]
似乎唯一的方法是：[这是一个糟糕的方法，不要这样做，见下面的注释]
if( '23' == (integer)'23' ) // return true
if( 23 == (integer)23 ) // return true
if( 23.3 == (integer)23.3 ) // return false
if( '23.3' == (integer)'23.3') // return false
But is there a function to do the above ?
但是有没有一个功能可以做到以上几点？
Just to clarify, I want the following results
只是为了澄清，我想要以下结果
23 // return true
'23' // return true
22.3 // return false
'23.3' // return false
Note: I just figured out my previous solution that I presented will return true for all strings. (thanks redreggae)
注意：我刚刚发现我之前提出的解决方案将对所有字符串返回 true。（感谢 redreggae）
$var = 'hello';
if( $var != (integer)$var )
echo 'not a integer';
// will return true! So this doesn't work either.
This is not a duplicate of Checking if a variable is an integer in PHP, because my requirements/definitions of integer is different than theres.
这不是检查变量是否是 PHP 中的整数的重复，因为我对整数的要求/定义与那里不同。
回答by bitWorking
try ctype_digit
if (!ctype_digit($_POST['id'])) {
// contains non numeric characters
}
Note: It will only work with stringtypes. So you have to cast to stringyour normal variables:
注意：它只适用于string类型。所以你必须转换到string你的正常变量：
$var = 42;
$is_digit = ctype_digit((string)$var);
Also note: It doesn't work with negative integers. If you need this you'll have to go with regex. I found thisfor example:
另请注意：它不适用于负整数。如果你需要这个，你将不得不使用正则表达式。我发现这个例如：
EDIT: Thanks to LajosVeres, I've added the D modifier. So 123\nis not valid.
编辑：感谢 LajosVeres，我添加了 D 修饰符。所以123\n是无效的。
if (preg_match("/^-?[1-9][0-9]*$/D", $_POST['id'])) {
echo 'String is a positive or negative integer.';
}
More: The simple test with casting will not work since "php" == 0 is trueand "0" === 0 is false!
See types comparisons tablefor that.
更多：带有转换的简单测试将不起作用，因为 "php" == 0 istrue并且 "0" === 0 is false！请参阅类型比较表。
$var = 'php';
var_dump($var != (int)$var); // false
$var = '0';
var_dump($var !== (int)$var); // true
回答by IT Vlogs
try filter_varfunction
尝试filter_var功能
filter_var($_POST['id'], FILTER_VALIDATE_INT);
use:
用：
if(filter_var($_POST['id'], FILTER_VALIDATE_INT)) {
//Doing somethings...
}
回答by Sébastien
In PHP $_POSTvalues are always text (stringtype).
在 PHP 中，$_POST值总是文本（string类型）。
You can force a variable into the integer type like this:
您可以将变量强制转换为整数类型，如下所示：
$int_id = (int)$_POST['id'];
That will work if you are certain that $_POST['id']should bean integer. But if you want to make absolutely sure that it contains only numbers from 0to 9and no other signs or symbols use:
如果您确定它$_POST['id']应该是整数，那将起作用。但是，如果您想绝对确保它只包含从0to 开始的数字9而没有其他符号或符号，请使用：
if( ctype_digit( $_POST['id'] ) )
{
$int_id = (int)$_POST['id'];
}
回答by siergiej
Check it out: http://php.net/manual/en/function.ctype-digit.php- it validates if string contains only digits, so be sure not to pass an int to that function as it will most likely return false; However all values coming from $_POSTare always strings so you are safe. Also it will not validate negative number such as -18since -is not a digit, but you can always do ctype_digit(ltrim($number, '-'))
检查一下：http: //php.net/manual/en/function.ctype-digit.php- 它验证字符串是否只包含数字，所以一定不要将 int 传递给该函数，因为它很可能会返回 false ; 但是，所有来自的值$_POST始终是字符串，因此您很安全。它也不会验证负数，例如-18因为-不是数字，但你总是可以这样做ctype_digit(ltrim($number, '-'))
is_intchecks the variable typewhich in your case is string; it would be the same as (integer)$v === $vas ==does some real obscure things in order to compare two variables of a different type; you should always use ===unless you want mess like "0af5gbd" == 0to return true
is_int检查您的情况下的变量类型string；这将是一样(integer)$v === $v的==呢，以比较不同类型的两个变量一些真正晦涩的东西; 你应该总是使用，===除非你想要像"0af5gbd" == 0返回 true一样混乱
Also, keep in mind that ctype_digitwill not tell you if the string can be actually converted to a valid intsince maximum integer value is PHP_INT_MAX; If your value is bigger than that, you will get PHP_INT_MAXanyway.
另外，请记住，它ctype_digit不会告诉您字符串是否可以实际转换为有效值，int因为最大整数值为PHP_INT_MAX; 如果你的价值大于那个，你PHP_INT_MAX无论如何都会得到。
回答by Lajos Veres
preg_match('/^\d+$/D',$variable) //edit
回答by CaM2091
Using is_numeric()for checking if a variable is an integer is a bad idea.
This function will send TRUEfor 3.14for example. It's not the expected behavior
使用is_numeric()检查，如果一个变量是一个整数是一个坏主意。此功能会发送TRUE用于3.14例如。这不是预期的行为
To do this correctly, you can use one of these options :
要正确执行此操作，您可以使用以下选项之一：
Considering this variables array :
考虑到这个变量数组：
$variables = [
"TEST 0" => 0,
"TEST 1" => 42,
"TEST 2" => 4.2,
"TEST 3" => .42,
"TEST 4" => 42.,
"TEST 5" => "42",
"TEST 6" => "a42",
"TEST 7" => "42a",
"TEST 8" => 0x24,
"TEST 9" => 1337e0
];
The first option (FILTER_VALIDATE_INT Way) :
第一个选项（FILTER_VALIDATE_INT 方式）：
# Check if your variable is an integer
if( ! filter_var($variable, FILTER_VALIDATE_INT) ){
echo "Your variable is not an integer";
}
Output :
输出 ：
TEST 0 : 0 (type:integer) is not an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?
The second option (CASTING COMPARISON Way) :
第二个选项（铸造比较方式）：
# Check if your variable is an integer
if ( strval($variable) != strval(intval($variable)) ) {
echo "Your variable is not an integer";
}
Output :
输出 ：
TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?
The third option (CTYPE_DIGIT Way) :
第三个选项（CTYPE_DIGIT方式）：
# Check if your variable is an integer
if( ! ctype_digit(strval($variable)) ){
echo "Your variable is not an integer";
}
Output :
输出 ：
TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?
The fourth option (REGEX Way) :
第四个选项（REGEX方式）：
# Check if your variable is an integer
if( ! preg_match('/^\d+$/', $variable) ){
echo "Your variable is not an integer";
}
Output :
输出 ：
TEST 0 : 0 (type:integer) is an integer ?
TEST 1 : 42 (type:integer) is an integer ?
TEST 2 : 4.2 (type:double) is not an integer ?
TEST 3 : 0.42 (type:double) is not an integer ?
TEST 4 : 42 (type:double) is an integer ?
TEST 5 : 42 (type:string) is an integer ?
TEST 6 : a42 (type:string) is not an integer ?
TEST 7 : 42a (type:string) is not an integer ?
TEST 8 : 36 (type:integer) is an integer ?
TEST 9 : 1337 (type:double) is an integer ?
回答by glerendegui
if you know it is a string variable (like post o get values), you can use:
如果你知道它是一个字符串变量（比如 post o get values），你可以使用：
function is_really_integer($var) {
return $var == (string)(integer)$var;
}
回答by David D
The accepted answer using ctype_digit is correct, however you can make life easier using a function. This will covert the variable to a string, so you don't have to:
使用 ctype_digit 接受的答案是正确的，但是您可以使用函数使生活更轻松。这会将变量转换为字符串，因此您不必：
function is_num($x){
if(!is_string($x)){
$x=(string)$x;
}
if(ctype_digit($x)){
return true;
}
return false;
}
Usage:
用法：
if (is_num(56)) {
// its a number
}
if (is_num('56')) {
// its a number
}
If you want to accept decimals too, use this:
如果您也想接受小数，请使用以下命令：
function is_num($x){
if(!is_string($x)){
$x=(string)$x;
}
if (strpos($x,'.')!==false) {
if(substr_count($x,'.')>1||strpos($x,'.')<1||strpos($x,'.')>=strlen($x)){
return false;
}
$x=str_replace('.','',$x);
}
if(ctype_digit($x)){
return true;
}
return false;
}
回答by Andreas
Use the following, universal function to check all types:
使用以下通用函数检查所有类型：
function is_digit($mixed) {
if(is_int($mixed)) {
return true;
} elseif(is_string($mixed)) {
return ctype_digit($mixed);
}
return false;
}
回答by Luca C.
I use:
我用：
is_int($val)||ctype_digit($val)
Note than this catch only positive integer strings
请注意，这仅捕获正整数字符串
