Change Row in MySQL using PHP Code - php

Question

I'm currently getting the error "Fatal Error: Cannot pass parameter 3 by reference in Update.php on line 14. I need to know how I can fix this.
Here's Update.php:
<?php
$post_json = file_get_contents("php://input");
$post = json_decode($post_json, true);
$response = array();
$configs = include('config.php');
$mysqli = new mysqli($configs["host"], $configs["username"], $configs["password"], "Account"); if($mysqli->connect_errno){
die("Error connecting to MySQL database (".$mysqli->connect_errno.") ".$mysqli->connect_error);
}
$name = $post["Name"];
$rank = $post["Rank"];
$perm = (int) $post["Perm"];
$stmt = $mysqli->prepare("UPDATE accounts SET `rank`=?,`rankPerm`=? WHERE `name`=?");
$stmt->bind_param("sis", $rank, (int) $perm, $name);
$stmt->execute();
?>
I believe that the error is in the (int) part of line 14, but I don't know what's wrong.
I'm trying to edit the "rank" part of one of these rows with this PHP file when executed:
http://imgur.com/a/Ms639
Thanks for all of your help in advance!
NOTE: Already checked update existing row in mysql using php coding

Answer 1

use this
<?php
$post_json = file_get_contents("php://input");
$post = json_decode($post_json, true);
$response = array();
$configs = include('config.php');
$mysqli = new mysqli($configs["host"], $configs["username"], $configs["password"], "Account"); if($mysqli->connect_errno){
die("Error connecting to MySQL database (".$mysqli->connect_errno.") ".$mysqli->connect_error);
}
$name = $post["Name"];
$rank = $post["Rank"];
$perm = (int) $post["Perm"];
$sis="sis";
$stmt = $mysqli->prepare("UPDATE accounts SET `rank`=?,`rankPerm`=? WHERE `name`=?");
$stmt->bind_param($sis, $rank, $perm, $name);
$stmt->execute();
?>
I believe you have to pass variables rather than a string.
Or you could use bindvalue() instead of bindparam() if you're using PDO.

Answer 2

From Php.net
Note:
Care must be taken when using mysqli_stmt_bind_param() in conjunction with call_user_func_array(). Note that mysqli_stmt_bind_param() requires parameters to be passed by reference, whereas call_user_func_array() can accept as a parameter a list of variables that can represent references or values.
It seems when you add a type conversion as a parameter to a function Php doesn't keep the reference, if you remove the (int) in the function call should be fine.
$stmt->bind_param("sis", $rank, (int) $perm, $name);
To
$stmt->bind_param("sis", $rank, $perm, $name);

Answer 3

May be this is the problem...
$name = $_post["Name"];
$rank = $_post["Rank"];
$perm = (int) $_post["Perm"];