Php Folks,

Look at this procedural style php prepared statement.
I want to know why this query fails:

"UPDATE users SET first_name = ?, middle_name = ?, surname = ?, gender = ?, marital_status = ?, working_status = ? WHERE id = ?";

Where is the syntax error ? Do you see any ? Cos I don't.
Prepare failed:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '"UPDATE users SET first_name = ?, middle_name = ?, surname = ?, gender = ?, mari' at line 1
"UPDATE users SET first_name = ?, middle_name = ?, surname = ?, gender = ?, marital_status = ?, working_status = ? WHERE id = ?";

Here is the complicated context:

//Connect to Database. (DB_SERVER, BD_USERNAME, DB_PASSWORD, DB_NAME).
$conn = mysqli_connect("localhost","root","","powerpage");
$conn->set_charset('utf8mb4'); //Always set Charset.
				
if($conn === false)
{
	die("ERROR: Connection Error!. " . mysqli_connect_error());
}
				
//Prepare an UPDATE Statement.
$sql_query = "\"UPDATE users SET ";//Half built the query. Now got to grab the Column Names (which need querying) that match the form Item Names.
foreach($form_questions_labels AS $form_question_label)
{
	$value = $form_question_label;
	$value_1 = str_replace(" ","_","$value"); //Replaced Spaces to Underscores.
	$value_2 = strtolower("$value_1"); //Replaced Upper Case to Lower Case.	
					
   if(isset($_POST["$value_2"]) && !empty($_POST["$value_2"]))
  {
      $sql_query = "$sql_query" . "$value_2 = ?, ";
  }
}

$sql_query = "$sql_query" . "WHERE id = ?\";";
//Now below, need to fix the last part of the Mysql Tbl Query as query has been generated something like this with a comma after the final column name's placeholder. Eg. $sql_query = "UPDATE users SET first_name = ?, surname = ?, WHERE id = ?";
$value_3 = $sql_query;
$value_4 = str_replace("= ?, WHERE","= ? WHERE","$value_3");//Replaced the Comma after the final column name's placeholder.
$sql_query = $value_4;
echo "$sql_query<br>";

$stmt = mysqli_prepare($conn,$sql_query);
if($stmt == false) 
{
    // just for debugging for now:
    die("<pre>Prepare failed:\n".mysqli_error($conn)."\n$sql_query</pre>");
}

Remember, when I echo the $sql_query, I see this:
"UPDATE users SET first_name = ?, middle_name = ?, surname = ?, gender = ?, marital_status = ?, working_status = ? WHERE id = ?";

Now where is the supposed syntax error there ?