PHP and MySQLi - Cannot pass parameter 2 by reference in [duplicate]
I am trying to make a function which will check update and insert some data but I am having an issue in the first step where the $stmt->bind_param is saying that is not passing parameters by reference or something like that.
I have attached below the function code:
public function killTarget($killerid,$victimiid,$victimcode)
{
if ($this->checkUsercode($victimcode,$victimiid))
{
$stmt = $this->_db->prepare("UPDATE users SET status =? WHERE user_id =?");
$stmt->bind_param("ii",0,$victimiid);
if ($stmt->execute())
{
$stmt->store_result();
$stmt->fetch();
$stmt = $this->_db->prepare("SELECT victim_id FROM target WHERE killer_id = ?");
$stmt->bind_param("i",$victimiid);
if ($stmt->execute())
{
$stmt->store_result();
$stmt->bind_result($targetid);
$stmt->fetch();
$stmt = $this->_db->prepare("INSERT INTO target (killer_id, victim_id) VALUES (?,?)");
$stmt->bind_param("ii",$killerid,$targetid);
if ($stmt->execute())
{
$stmt->store_result();
$stmt->fetch();
$stmt->close();
}
}
}
else
{
Main::setMessage("targets.php",$this->_db->error,"alert-error");
}
}
}
Solution 1:
You cannot do this in mysqli:
$stmt->bind_param("ii",0,$victimiid);
The 0
needs to be a variable.
Try this:
$zero = 0;
$stmt->bind_param("ii",$zero,$victimiid);
Solution 2:
Watch out! mysqli_stmt::bind_param
accepts a reference to a variable, not a constant value. Therefore you have to create a variable to hold that 0
and then reference that variable instead.
$i = 0;
$stmt->bind_param("ii", $i, $victimiid);
Solution 3:
Make 0 a variable or include it directly in your query.
$zero = 0;
$stmt->bind_param("ii", $zero, $victimiid);