I have the following PHP:
<?php
$array = array("1","2","3");
$only_integers === array_filter($array,'is_numeric'); // true
if($only_integers == TRUE)
{
echo 'right';
}
?>
For some reason it always returns nothing. I don't know what I'm doing wrong.
Thanks
is_int checks the actual type of a variable, which is string in your case. Use is_numeric for numeric values regardless of variable type.
Note that the following values are all considered "numeric":
"1"
1
1.5
"1.5"
"0xf"
"1e4"
i.e. any floats, integers or strings that would be valid representations of floats or integers.
Edit: Also, you might have misunderstood array_filter, it does not return true or false but a new array with all values for which the callback function returned true. if($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.
is_int, is_numeric really isn't that great as an alternative, because it returns true for a lot of values that you wouldn't necessarily want (hex, exponents, etc). You might want to use ctype_digit() instead.
ctype_digit is a good alternative for only integers in decimal format, thanks for adding it.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.
if(count($only_integers) == count($array)) {
echo 'right';
} else {
echo 'wrong';
}
=== $array part of the accepted answer there.
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.
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.
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.
$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($only_integers) > 0)
{
echo 'right';
}
count(): there are many ways of doing the same thing. To explain his code: In the any() function, he is comparing it with an empty array; if it equals an empty array, then we know that all the items were filtered; none of them matched, so the function can therefore return false. In the all() function, he is comparing it with the original input; if the filtered array is different from the original, then we know that at least one element has been filtered, so the function can return false.all() type check or an any() type check. My answer is an any() check -- ie it gives success if any of the values are good. If you want all(), change the count()` check to if($only_integers === $array), as per the answer on the other page.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($only_integers == TRUE)
{
echo 'right';
}
?>
array_filter($array,'is_int'); no, not really! a float is considered to be a numeric value as well as a string like "34" therefore the is_numeric the author posted was right.
is_int instead of is_numberic like the other question has mentioned :)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.
<?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.
