CODEWITHSUNDEEP
phparraysfunctionintegernumeric
Michael Samuel
Michael Samuel

Reputation: 3920

PHP code to check if array is numeric is not working

I have the following PHP:

 <?php

 $array = array("1","2","3");
 $only_integers === array_filter($array,'is_numeric'); // true

 if($only_integers == TRUE)
 {
 echo 'right';
 }

 ?>

For some reason it always returns nothing. I don't know what I'm doing wrong.

Thanks

Upvotes: 1

Views: 2761

Answers (6)

Tzu ng
Tzu ng

Reputation: 9244

Do you try to run your code before posting ? I have this error : 

Notice: Undefined variable: only_integers in ~/php/test.php on line 4
Notice: Undefined variable: only_integers in ~/php/test.php on line 6

Change === to = fixes the problem right away. You better learn how to use phplint and other tools to avoid typo mistake like this.

Upvotes: 0

SDC
SDC

Reputation: 14222

  1. is_int() will return false for "1" because it's a string.
    I see you've edited the question now to use is_numeric() instead; this is also probably a bad idea, as it will return true for hex and exponent values, which you probably don't want (eg is_numeric("dead") will return true).
    I suggest using ctype_digit() instead.

  2. The triple-equal is being misused here. It is used for a comparison, not an assignment, so $only_integers will never be set. Use single-equal to set $only_integers.

  3. array_filter() doesn't return a true/false value; it returns the array, with the filtered values removed. This means that the subsequent check that $only_integers is true will not work.

  4. $only_integers == TRUE. This is okay, but you probably should have used the triple-equal here. But of course, we already know that $only_integers won't be true or false, it'll be an array, so actually we need to check whether it's got any elements in it. count() would do the trick here.

Here's what your code looks like, taking all that into account...

 $array = array("1","2","3");
 $only_integers = array_filter($array,'ctype_digit'); // true

 if(count($only_integers) > 0)
 {
     echo 'right';
 }

Upvotes: 1

Matthew Purdon
Matthew Purdon

Reputation: 773

You have to compare the length of the original array to the length of the filtered array. The array_filter function returns an array with values matching the filter set to true.

http://php.net/array_filter

 if(count($only_integers) == count($array))  {
     echo 'right';
 } else {
     echo 'wrong';
 }

Upvotes: 2

Fabian Schmengler
Fabian Schmengler

Reputation: 24576

is_int checks the actual type of a variable, which is string in your case. Use is_numeric for numeric values regardless of variable type.

Note that the following values are all considered "numeric":

"1"
1 
1.5
"1.5"
"0xf"
"1e4"

i.e. any floats, integers or strings that would be valid representations of floats or integers.

Edit: Also, you might have misunderstood array_filter, it does not return true or false but a new array with all values for which the callback function returned true. if($only_integers) works nonetheless (after you fixed your assignment operator) because all non-empty arrays are considered "true-ish".

Edit 2: as @SDC pointed out, you should use ctype_digit if you only want to allow integers in decimal format.

Upvotes: 2

Embedded Software Engineer
Embedded Software Engineer

Reputation: 4904

change === with = it used to compare not for initialise a variable

<?php

 $array = array(1,2,3);
 $only_integers = array_filter($array,'is_int'); // true

 if($only_integers == TRUE)
 {
 echo 'right';
 }

?>

Upvotes: 0

Can Geliş
Can Geliş

Reputation: 1474

<?php
$test1 = "1";
if (is_int($test1) == TRUE) {
    echo '$test1 is an integer';
}
$test2 = 1;
if (is_int($test2) == TRUE) {
    echo '$test2 is an integer';
}
?>

try this code and you'll understand why your code doesn't work.

Upvotes: -1

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