如何检查从文本输入返回的字符串值是否为整数

I have a form where users search for keywords.

I want to check if users submit a number (integer) instead of a keyword.

I use this:

$keyword = $_REQUEST["keyword"];

if (is_int($keyword)) {
  echo 'true';
} else {
  echo 'false';
}

but it always returns false even when the value submitted is integer.

is_int checks to see whether the variable is of integer type. What you have is a string whose characters happen to represent an integer.

Try:

if (is_numeric($keyword) && is_int(0+$keyword))

The 0+ will implicitly convert the following to a number type, so you'll end up with the numeric value of the string if possible. Calling is_numeric first ensures you aren't converting garbage.

is_int() is for type safe comparison, so the string that is received from the browser always evaluates to false.

If you really need to check if a variable is an integer you can use this handy function:

function str_is_int($number) {
    // compare explicit conversion to original and check for maximum INT size
    return ( (int)$number == $number && $number <= PHP_INT_MAX );  
    // will return false for "123,4" or "123.20"
    // will return true for "123", "-123", "-123,0" or "123.0"
}

Note: you should replace PHP_INT_MAX with the maximum integer size of your target system - e.g. your database.

in the docs of is_int() you can read this:

is_int — Find whether the type of a variable is integer

bool is_int ( mixed $var )

Note:

To test if a variable is a number or a numeric string (such as form input, which is always a string), you must use is_numeric().

Example:

<?php
if (is_int(23)) {
    echo "is integer
";
} else {
    echo "is not an integer
";
}
var_dump(is_int(23));
var_dump(is_int("23"));
var_dump(is_int(23.5));
var_dump(is_int(true));
?>

The above example will output:

is integer
bool(true)
bool(false)
bool(false)
bool(false)

I think ctype_digit() will work in your case.