Why am i unable to insert a data in to data base - mysql

Question

These are the codes for the page:
<?php
session_start();
if(isset($_SESSION['level'])){
if($_SESSION['level'] == 2 ){
require("../db/dbConn.php");
$submitted = isset($_POST['submit']);
if($submitted){
//check user's input
if(isset($_POST['issue_type'])){
$issue_type =$_POST['issue_type'];
}
else {
$issue_type = null;
echo '<p><font color="red">Please Select a Issue Type</font></p>';
}
if(isset($_POST['description'])){
$description=$_POST['description'];
}
else{
$description = null;
echo '<p><font color="red">You forgot to enter a description</font> </p>';
}
if(isset($_POST['reported_account_id'])){
$reported_account_id = $_POST['reported_account_id'];
}
else{
$reported_account_id = null;
echo '<p><font color="red">You forgot to enter your ID</font></p>';
}
if(isset($_POST['DateTimeCreated'])){
$DateTimeCreated=$_POST['DateTimeCreated'];
}
else{
$DateTimeCreated= null;
echo '<p><font color="red">You forgot to enter the date and time of the Issue </font></p>';
}
//Prepare the Insert Statement
$stmt = "INSERT INTO problem (issue_id, description, reported_account_id, DateTimeCreated) VALUES ('$issue_type', '$description','$reported_account_id','$DateTimeCreated')";
$result = mysqli_query($conn, $stmt);
$conn->close();
//TODO 5: Check result of executing insert statement and rows inserted. Print user's input if 1 row is inserted successfully,
// else print error message
if($result==true){
echo '<p><font color="green">The problem has been created. Thank you</font></p>';
echo '<p>Registration Successful Please Click Here';
} else {
echo "<p><font color=red><b>Data not saved. Please try again</b></font></p>";
echo '<p>Inserting Failed Please Click Here to Try Again';
}
}}
} else {
header("Location: ../index.php");
}
?>
I am not able to submit the insert the details from the form page into the database. That is the only issue that i am faced with. Please help me point out the errors that i made.

Answer 1

To Fix the issue, please replace
$stmt = "INSERT INTO problem (issue_id, description, reported_account_id, DateTimeCreated) VALUES ('$issue_type', '$description','$reported_account_id','$DateTimeCreated')";
With
$stmt = "INSERT INTO problem (issue_id, description, reported_account_id, DateTimeCreated) VALUES ('"+$issue_type+"', '"+$description+"','"+$reported_account_id+"','"+$DateTimeCreated+"')";