Convert implode string to interger variable PHP

Question

Suppose after applying implode to an array , I got a String for example :-

$var = 1631075,1631076;

On applying var_dump to $var , I received the output as string(15) "1631075,1631076"

how will I convert entire $var into Integer variable . Such that the var_dump display int(15)1631075,1631076

That doesn't look like a valid integer. Do you want to get the integer 16310751631076? — Mureinik
– Mureinik, Commented Jun 8, 2018 at 17:17
I dont understand your question, do you want to create an array of integer? — Azhy
– Azhy, Commented Jun 8, 2018 at 17:18
A valid integer will not include and commas only numbers. You format the integer->string to show commas for easy readability later. Strip out any non-numbers and do $var = intval($string_with_only_numbers); — Danny
– Danny, Commented Jun 8, 2018 at 17:21
Plus, that comma is not in the correct place as a thousands separator. — AbraCadaver
– AbraCadaver, Commented Jun 8, 2018 at 17:24

Danny · Accepted Answer · 2018-06-08 17:30:55Z

1

$var = '1631075,1631076';
foreach(explode(',',$var) as $val){
 echo intval($val);
}

or something like

$var = '1631075,1631076';
$integers = explode(',',$var);
$converted = array_map('intval', $integers);

answered Jun 8, 2018 at 17:30

Danny

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4 Comments

But Sir ,it is giving the output as 16310751631076; how can I get 1631075,1631076 as integer

"1631075,1631076" is not an integer and can't be. That can only be a string if you want that inside one variable.

Sir actually I want the final output as [1631075,1631076] (integer format) the required array is Array ( [0] => 1631075 [1] => 1631076 )

Use "intval()" on the array values before handling mathematical functions. Also look at stackoverflow.com/questions/8963910/…