Check if any number in the array is not a int

Question

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.

diggersworld · Answer 1 · 2013-02-16T18:48:56.563

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.

score 0 · Accepted Answer · 2013-02-16T18:54:22.223

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>';
 }
}

$number is just the user input you used for calculating average!!! which not correct — , Feb 16 '13 at 18:59

score 0 · Answer 3 · answered Feb 16 '13 at 18:54

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;
     }

Check if any number in the array is not a int

3 Answers3