Object of class mysqli could not be converted to string - php

Question

I had my old website so I've decided to change it from mysql_ to mysqli so I've managed to complete 40% and now i am stuck with this problem.Help Me!
I am getting error on 'implode()' function
function user_data($user_id,$conn){
$data = array();
$user_id = (int)$user_id;
$func_num_args = func_num_args();
$func_get_args = func_get_args();
if ($func_num_args > 1){
unset($func_get_args[0]);
$fields = '`'.implode('`,`',$func_get_args).'`';
$query = "SELECT ".$fields." FROM users WHERE user_id = ".$user_id."";
$result = mysqli_query($conn,$query) or die(mysqli_error($conn));
while ($row = $result->fetch_assoc()) {
$data = $row['user_id'];
}
return $data;
}
}
In order to get below code to work properly
if (logged_in() === true){
$session_user_id = $_SESSION['user_id'];
$user_data = user_data($session_user_id,'user_id','username',
'password','first_name','last_name','email','type',$conn);
}
Any alternate way to perform same task will

Answer 1

You have not correct definition of user_data function,
In it's signature you have only two arguments:
function user_data($user_id, $conn)
So, these arguments are $user_id and $conn.
But when you call your user_data you pass more than 2 arguments:
user_data($session_user_id,'user_id','username', 'password','first_name','last_name','email','type',$conn);
See, you have 8 arguments here. And $conn is not the second one, it's eighth!
And when you do
$fields = '`'.implode('`,`',$func_get_args).'`';
last argument which holds your mysqli-connection is being added to $fields.
So, you have to rewrite your function, for example this way:
function user_data($user_id, $conn, $fields) {
$data = array();
$user_id = (int)$user_id;
$fields = '`'.implode('`,`', $fields).'`';
$query = "SELECT ".$fields." FROM users WHERE user_id = ".$user_id."";
$result = mysqli_query($conn, $query) or die(mysqli_error($conn));
while ($row = $result->fetch_assoc()) {
$data = $row['user_id'];
}
return $data;
}
And call it for example:
$user_data = user_data(
$session_user_id, // $user_id
$conn, // $conn
array('user_id','username','password','first_name','last_name','email','type') // fields as ARRAY
);

Answer 2

$func_get_args has $user_id, other string values and at last the mysqli connection object. You must unset last element of function parameters. Correct user_data function is that:
function user_data($user_id,$conn){
$data = array();
$user_id = (int)$user_id;
$func_num_args = func_num_args();
$func_get_args = func_get_args();
if ($func_num_args > 1){
unset($func_get_args[0]);
unset($func_get_args[ $func_num_args - 1]); // you must delete last element becouse this is mysqli object
$fields = '`'.implode('`,`',$func_get_args).'`';
$query = "SELECT ".$fields." FROM users WHERE user_id = ".$user_id."";
$result = mysqli_query($conn,$query) or die(mysqli_error($conn));
while ($row = $result->fetch_assoc()) {
$data = $row['user_id'];
}
return $data;
}
}