how to insert values with comma?in switch statement - php

Question

i am using switch statements to insert comma values:
<?php
session_start();
include('config1.php');
$category_id = 1;
$AnswerID = $_POST['AnswerID'];
$questionid = $_POST['questionid'];
$timetaken = $_POST['timetaken'];
$limit = $_POST['limit'];
echo "$limit";
$bd = "$limit";
switch ($bd) {
case"1":
$sql = "INSERT INTO results (id, user_id, category_id, q_id, answer_id, time_taken)
VALUES (',', '".$_SESSION['id']."', '$category_id', '$questionid', '$AnswerID', '$timetaken')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$last_id = mysqli_insert_id($conn);
echo "Last inserted ID is: " . $last_id;
// Set session variables
$_SESSION["last_id"] = "$last_id";
break;
case"2":
quiz_test();
break;
case"3":
quiz_test();
break;
case"4":
quiz_test();
break;
case"5":
quiz_test();
unset($_SESSION['last_id']);
break;
default:
echo "something is wrong";
}
function quiz_test(){
$sql = "SELECT q_id, answer_id, time_taken FROM results WHERE id='" . $_SESSION["last_id"] . "'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$qid=$rows['q_id'];
$ans=$rows['answer_id'];
$time=$rows['time_taken'];
}
}
$conn->query("update results set q_id =('$questionid','$qid'),answer_id = ('$AnswerID','$ans'),time_taken=('$timetaken','$time') where id = '" . $_SESSION["last_id"] . "'");
}
?>
In case 1 insert values and get insert id,and set into session.
Case 2 select,update statement are not working.i got following errors:
Notice: Undefined variable: conn in C:\xampp\htdocs\N\exam\exam\DOCS\Insert.php on line 62
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\N\exam\exam\DOCS\Insert.php on line 62
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\N\exam\exam\DOCS\Insert.php on line 63
Notice: Undefined variable: conn in C:\xampp\htdocs\N\exam\exam\DOCS\Insert.php on line 70
Fatal error: Call to a member function query() on a non-object in C:\xampp\htdocs\N\exam\exam\DOCS\Insert.php on line 70

Answer 1

You need to add one parameter $conn in your function quiz_test() then it will works
Try below code
<?php
session_start();
?>
<?php
include('config1.php');
$category_id = 1;
$AnswerID = $_POST['AnswerID'];
$questionid = $_POST['questionid'];
$timetaken = $_POST['timetaken'];
$limit = $_POST['limit'];
echo "$limit";
$bd = "$limit";
switch ($bd) {
case"1":
$sql = "INSERT INTO results (id, user_id, category_id, q_id, answer_id, time_taken)
VALUES (',', '".$_SESSION['id']."', '$category_id', '$questionid', '$AnswerID', '$timetaken')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$last_id = mysqli_insert_id($conn);
echo "Last inserted ID is: " . $last_id;
// Set session variables
$_SESSION["last_id"] = "$last_id";
break;
case"2":
quiz_test($conn);
break;
case"3":
quiz_test($conn);
break;
case"4":
quiz_test($conn);
break;
case"5":
quiz_test($conn);
unset($_SESSION['last_id']);
break;
default:
echo "something is wrong";
}
function quiz_test($conn){
$sql = "SELECT q_id, answer_id, time_taken FROM results WHERE id='" . $_SESSION["last_id"] . "'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$qid=$rows['q_id'];
$ans=$rows['answer_id'];
$time=$rows['time_taken'];
}
}
$conn->query("update results set q_id =('$questionid','$qid'),answer_id = ('$AnswerID','$ans'),time_taken=('$timetaken','$time') where id = '" . $_SESSION["last_id"] . "'");
}
?>

Answer 2

You appear to be mixing together two different flavors of PHP API functions. You also, in your call to mysqli_insert_id() you should not be specifying the id if it is an autoincrement column. I believe you intended to do something along these lines:
$sql = "INSERT INTO results (user_id, category_id, q_id, answer_id, time_taken)
VALUES ('".$_SESSION['id']."', '$category_id', '$questionid', '$AnswerID', '$timetaken')";
mysqli_query($conn, $sql);
$last_id = mysqli_insert_id($conn);
echo "Last inserted ID is: " . $last_id;
And the other problem as #Krish pointed out is that you need to pass your connection variable $conn to the quiz_test() function, e.g.
function quiz_test($conn) {
...
}