Check if any number in the array is not a int [duplicate] - php

Question

This question already has answers here:
More concise way to check to see if an array contains only numbers (integers)
(6 answers)
Closed 10 years ago.
I made a form that you can send numbers with for example
1 2 3 4 5 6 7 8 9 10
And then I made a PHP code that will grab these numbers inside a array & loop that will calculate how many array items there is and then do the calculations.
I used is_numeric to check if the array contains only ints.
But for some reason it doesn't really work.
<?php
$total = null;
$number = $_POST['number'];
$numbers = explode(" ", $number);
foreach($numbers as $number) {
$total = $total + $number;
}
$notnumber = '<center>You must enter a number</center>';
$empty = '<center>The field is empty.</center>';
if ($numbers == is_numeric($numbers) && $total != null) {
$avg = $total / $number;
echo '<center>Avarge is: <b>'.$avg.'</b></center>';
} else if ($_POST['number'] == "") {
echo $empty;
} else if ($numbers != is_numeric($numbers)) {
echo $notnumber;
}
?>
This is the form
<form action="index.php" method="post">
<input type="text" name="number" class="input"><br /><br />
<input type="submit" value="Calculate results">
</form>
What happens:
When I enter numbers, it will echo the error "$notnumbers" yet they are numbers.
What have I done wrong?
Thanks.

Answer 1

if($numbers == is_numeric($numbers) ... is not correct
use
if (is_numeric($numbers) && $total != null) {
The better function is ctype_digit
Please try this and do not use yours!
//$_POST['number'] = '1 2 3 4 5 6 7 8 9 10';
$notnumber = '<center>You must enter a number</center>';
$empty = '<center>The field is empty.</center>';
if(!isset($_POST['number']) || strlen($_POST['number']) <= 0){
echo $empty;
}
else{
if( !preg_match('/^([1-9]{1})([0-9]*)(\s)([0-9]+)/', $_POST['number']) ){
echo $notnumber;
}else{
$total = null;
$number = $_POST['number'];
$numbers = explode(" ", $number);
//remove after test
echo "<pre>";
print_r($numbers);
echo "</pre>";
//end remove
$total = array_sum($numbers);
//count(array) = number of elements inside $array
$avg = $total / count($numbers);
echo '<center>Avarge is: <b>'.$avg.'</b></center>';
}
}

Answer 2

This is incorrect: $numbers == is_numeric($numbers).
Alternate Solution:
You could try this, providing you count 0 as an invalid number.
$number = $_POST['number'];
$numbers = explode(" ", $number);
$total = 0;
foreach($numbers as &$number) {
// covert value to integer
// any non-numerics become 0
$number = (int) $number;
$total += $number;
}
Now you won't have to worry about checking for is_numeric or is_int because they'll all be ints. You'll just need to check for 0 values instead.
Extra Tip:
<center> tag is also deprecated in HTML.CSS should be used to display centred text.

Answer 3

try this i solve some problem on code
<?php
$total = NULL;
$number = #$_POST['number'];
$numbers = explode(" ", $number);
$Num=0;
foreach($numbers as $number) {
$total = $total + $number;
$Num++;
}
$notnumber = '<center>You must enter a number</center>';
$empty = '<center>The field is empty.</center>';
if (is_array($numbers) && $total != NULL) {
$avg = $total / $Num;
echo '<center>Avarge is: <b>'.$avg.'</b></center>';
} else if (!isset($_POST['number'])) {
echo $empty;
} else if (!is_numeric($number)) {
echo $notnumber;
}