How to check if an integer is within a range of numbers in PHP?

问题

How can I check if a given number is within a range of numbers?

回答1:

The expression:

 ($min <= $value) && ($value <= $max)

will be true if $value is between $min and $max, inclusively

See the PHP docs for more on comparison operators

回答2:

You can use filter_var

filter_var(
    $yourInteger, 
    FILTER_VALIDATE_INT, 
    array(
        'options' => array(
            'min_range' => $min, 
            'max_range' => $max
        )
    )
);

This will also allow you to specify whether you want to allow octal and hex notation of integers. Note that the function is type-safe. 5.5 is not an integer but a float and will not validate.

Detailed tutorial about filtering data with PHP:

• https://phpro.org/tutorials/Filtering-Data-with-PHP.html

回答3:

Might help:

if ( in_array(2, range(1,7)) ) {
    echo 'Number 2 is in range 1-7';
}

http://php.net/manual/en/function.range.php

回答4:

You could whip up a little helper function to do this:

/**
 * Determines if $number is between $min and $max
 *
 * @param  integer  $number     The number to test
 * @param  integer  $min        The minimum value in the range
 * @param  integer  $max        The maximum value in the range
 * @param  boolean  $inclusive  Whether the range should be inclusive or not
 * @return boolean              Whether the number was in the range
 */
function in_range($number, $min, $max, $inclusive = FALSE)
{
    if (is_int($number) && is_int($min) && is_int($max))
    {
        return $inclusive
            ? ($number >= $min && $number <= $max)
            : ($number > $min && $number < $max) ;
    }

    return FALSE;
}

And you would use it like so:

var_dump(in_range(5, 0, 10));        // TRUE
var_dump(in_range(1, 0, 1));         // FALSE
var_dump(in_range(1, 0, 1, TRUE));   // TRUE
var_dump(in_range(11, 0, 10, TRUE)); // FALSE

// etc...

回答5:

if (($num >= $lower_boundary) && ($num <= $upper_boundary)) {

You may want to adjust the comparison operators if you want the boundary values not to be valid.

回答6:

You can try the following one-statement:

if (($x-$min)*($x-$max) < 0)

or:

if (max(min($x, $max), $min) == $x)

回答7:

using a switch case

    switch ($num){

        case ($num>= $value1 && $num<= $value2): 
            echo "within range 1";
        break;
        case ($num>= $value3 && $num<= $value4): 
            echo "within range 2";
        break;
        .
        .
        .
        .
        .

        default: //default
            echo "within no range";
        break;
     }

回答8:

Another way to do this with simple if/else range. For ex:

$watermarkSize = 0;

if (($originalImageWidth >= 0) && ($originalImageWidth <= 640)) {
    $watermarkSize = 10;
} else if (($originalImageWidth >= 641) && ($originalImageWidth <= 1024)) {
    $watermarkSize = 25;
} else if (($originalImageWidth >= 1025) && ($originalImageWidth <= 2048)) {
    $watermarkSize = 50;
} else if (($originalImageWidth >= 2049) && ($originalImageWidth <= 4096)) {
    $watermarkSize = 100;
} else {
    $watermarkSize = 200;
}

回答9:

Some other possibilities:

if (in_array($value, range($min, $max), true)) {
    echo "You can be sure that $min <= $value <= $max";
}

Or:

if ($value === min(max($value, $min), $max)) {
    echo "You can be sure that $min <= $value <= $max";
}

Actually this is what is use to cast a value which is out of the range to the closest end of it.

$value = min(max($value, $min), $max);

Example

/**
 * This is un-sanitized user input.
 */
$posts_per_page = 999;

/**
 * Sanitize $posts_per_page.
 */
$posts_per_page = min(max($posts_per_page, 5), 30);

/**
 * Use.
 */
var_dump($posts_per_page); // Output: int(30)

回答10:

I created a function to check if times in an array overlap somehow:

    /**
     * Function to check if there are overlapping times in an array of \DateTime objects.
     *
     * @param $ranges
     *
     * @return \DateTime[]|bool
     */
    public function timesOverlap($ranges) {
        foreach ($ranges as $k1 => $t1) {
            foreach ($ranges as $k2 => $t2) {
                if ($k1 != $k2) {
                    /* @var \DateTime[] $t1 */
                    /* @var \DateTime[] $t2 */
                    $a = $t1[0]->getTimestamp();
                    $b = $t1[1]->getTimestamp();
                    $c = $t2[0]->getTimestamp();
                    $d = $t2[1]->getTimestamp();

                    if (($c >= $a && $c <= $b) || $d >= $a && $d <= $b) {
                        return true;
                    }
                }
            }
        }

        return false;
    }

回答11:

Here is my little contribution:

function inRange($number) {
  $ranges = [0, 13, 17, 24, 34, 44, 54, 65, 200];
  $n = count($ranges);

  while($n--){
    if( $number > $ranges[$n] )
      return $ranges[$n]+1 .'-'. $ranges[$n + 1];
  }

回答12:

function limit_range($num, $min, $max)
{
  // Now limit it
  return $num>$max?$max:$num<$min?$min:$num;
}

$min = 0;  // Minimum number can be
$max = 4;  // Maximum number can be
$num = 10;  // Your number
// Number returned is limited to be minimum 0 and maximum 4
echo limit_range($num, $min, $max); // return 4
$num = 2;
echo limit_range($num, $min, $max); // return 2
$num = -1;
echo limit_range($num, $min, $max); // return 0

来源：https://stackoverflow.com/questions/4684023/how-to-check-if-an-integer-is-within-a-range-of-numbers-in-php