PHP function to check if the number belongs to the interval - php

Question

There is an array:
$bounds = array([0]=>array('lower'=>2,'upper'=>5),
[1]=>array('lower'=>0,'upper'=>3));
and a variable:
$val = 4;
Is there any PHP function that can say whether $val belongs to any interval defined by 'lower' and 'upper' bounds in $bounds array? In this example 4 belongs to the 1st interval [2; 5]. So, the answer should be 'true'.

Answer 1

I don't think there is a built-in function to do this.
However, you can do it with a foreach statement:
function check_interval($bounds, $val) {
foreach ($bounds as $array) {
if($array['lower'] <= $val && $array['upper'] >= $val)
return true;
}
return false;
}

Answer 2

I'm not aware of any. You'll probably have to code it. Something like this will do:
function isFromInterval($bounds, $val) {
foreach ($bounds as $value) {
if ($val >= $value['lower'] && $val <= $value['upper']) {
return true;
}
}
return false;
}

Answer 3

No.
You would have to make a loop for the array like this
$val = 4;
$key_id = FALSE;
foreach($bounds as $key => $data){
if($val <= $data['upper'] AND $val >= $data['lower']){
$key_id = $key;
break;
}
}
if($key_id !== FALSE){
// found something
// $bounds[$key_id] is your result in the array
} else {
// found nothing
}
As a function
function find_range($bounds=array(), $val=0, $return_key=TRUE){
if(is_array($bounds) === FALSE){
$bounds = array();
}
if(is_numeric($val) === FALSE){
$val = 0;
}
if(is_bool($return_key) === FALSE){
$return_key = TRUE;
}
$key_id = FALSE;
foreach($bounds as $key => $data){
if($val < $data['upper'] AND $val > $data['lower']){
$key_id = $key;
break;
}
}
if($key_id !== FALSE){
return ($return_key === TRUE ? $key_id : TRUE);
} else {
return FALSE;
}
}

Answer 4

No, but you can do:
$bounds = array(3=>array('lower'=>2,'upper'=>5),
4=>array('lower'=>0,'upper'=>3));
$val = 4;
foreach($bounds as $num => $bound){
if(max($bound) >= $val && $val >= min($bound)){
echo $num;
}
}