Convert json string to integer in php

Question

I'm using xampp and even though I've set the id column as integer in my table, the output is still string i.e I get "id":"1" instead of "id":1. I've come across a possible solution via JSON_NUMERIC_CHECK but I don't know how to implement this in my php script. Can someone show me how to modify my php script in order to have the id output as an integer.

    <?php


    require("config.inc.php");
    $query_params=null;



    $query = "Select * FROM feeds";


    try {
        $stmt   = $db->prepare($query);
        $result = $stmt->execute($query_params);
    }
    catch (PDOException $ex) {
        $response["success"] = 0;
        $response["message"] = "Database Error!";
        die(json_encode($response));
    }


    $rows = $stmt->fetchAll();


    if ($rows) {
        $response["feed"]   = array();

        foreach ($rows as $row) {
            $post             = array();
            $post["id"] = $row["id"];
            $post["name"]    = $row["name"];
            $post["image"]  = $row["image"];

            array_push($response["feed"], $post);
        }


        echo json_encode($response);


    } else {
        $response["success"] = 0;
        $response["message"] = "No Post Available!";
        die(json_encode($response));
    }

?>

If you have PHP 5.3.3 then just json_encode($response, JSON_NUMERIC_CHECK); — AbraCadaver
– AbraCadaver, Commented Feb 23, 2015 at 21:06

Jay Blanchard · Accepted Answer · 2015-02-23 21:06:49Z

8

Just add the option to the json_encode() call -

json_encode( $response, JSON_NUMERIC_CHECK );

answered Feb 23, 2015 at 21:06

Jay Blanchard

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Marc B · Accepted Answer · 2015-02-23 21:07:36Z

3

json_encode() goes off whatever PHP says the value's type is:

php > $arr = array('id' => '1');
php > var_dump($arr);
array(1) {
  ["id"]=>
  string(1) "1"
}
php > echo json_encode($arr);
{"id":"1"}

Since your 1 is a string in PHP, it'll be a string in JSON as well. So force it to be an int:

php > $arr = array('id' => (int)'1');
                           ^^^^^-----note the typecast here.
php > var_dump($arr);
array(1) {
  ["id"]=>
  int(1)
}
php > echo json_encode($arr);
{"id":1}

answered Feb 23, 2015 at 21:07

Marc B

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