A seemingly minor problem are bothering me, something that should have a simple solution. I am currently running a server and a database, the database which contains data that I want to use in a PHP request on a website. I have created a function in PHP that does the following:

function getUserID($val){
     include("config.php");
     $conn = new mysqli($dbservername, $dbusername, $dbpassword, $dbname) or die($conn);
     $sql = "SELECT userid FROM users WHERE username=?";
     $stmt = $conn->prepare($sql);
     $stmt->bind_param("s", $val);
     $stmt->execute();
     $result = $stmt->get_result();
     if(getUserExist($val)){
          $rowdata = mysqli_fetch_assoc($result);
          $conn->close();
          return $rowdata['userid'];
     }
}

This works just fine.. HOWEVER. The returned data type, which is supposed to be an Integer ( 1, 2, 3, 4... etc. ), returns a value similar to an JSON object or int(1), depending on how I write it.

array(1) { ["userid"]=> int(4) }

I have tried:

1. $rowdata['userid']
2. $rowdata

How do I make the function return purely the integer value? When it is added to the database with following code-snippet:

...
        $stmt = $conn->prepare("INSERT INTO users (user1, user2, user3) VALUES (?, ?, ?)");
        $stmt->bind_param("isi", $user1, $user2, $user3);
        $user1 = $_POST[getUserID($username)];
        $user2 = $_POST['val2'];
        $user3 = $_POST['val3'];
        $stmt->execute();
        
        $stmt->close();
    }
    $conn->close();

As mentioned, it retrieves data just fine, it is just the value that acts in an odd way. When this code is executed, $user1 or rather the final value within database has the value of NULL. (database accepts only integers in that slot).

CodePudding user response：

It looks to me as if the problem does not lie in getUserID(), but in this line:

$user1 = $_POST[getUserID($username)];

What you're doing here is not setting $user1 to the value of getUserID() - instead, you're setting it to "the element in the $_POST array which has a key of whatever getUserID() returns". And there are very few scenarios where that makes sense.

I'm assuming the line you want to replace it with is

$user1 = getUserID($username);