Notice: undefined variable

Question

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I dont know how to change the code to define the variable in php
i am trying to do form html to insert data into mysql database.

What I have tried:

<?php
$server = "localhost";
$username ="root";
$password = "";
$dbname = "booking";

$conn = mysqli_connect($server, $username, $password, $dbname);

if(isset($_POST['submit'])){
	 if(!empty($_POST['usertype']) && !empty($_POST['userid']) && !empty($_POST['usertype']) && !empty($_POST['bookingdate']) && !empty($_POST['startingtime']) && !empty($_POST['facility']) && !empty($_POST['payment']) )
$usertype=$_POST['usertype'];
$userid=$_POST['userid'];
$bookingdate=$_POST['bookingdate'];
$startingtime=$_POST['startingtime'];
$facility=$_POST['facility'];
$payment=$_POST['payment'];
}
$sql = "INSERT INTO booking(B_UserType,B_UserID, B_Date,B_StartTime,B_Facility ,B_PaymentType) VALUES ('$usertype','$userid','$bookingdate','$startingtime','$facility','$payment')";
  $run = mysqli_query($conn, $sql ) or die (mysqli_error());
if($run)
{
	echo "Inserted";
}else
{
	echo "Not inserted";
}
?>

Posted 20-Jul-20 21:59pm

steph99999

Updated 20-Jul-20 23:09pm

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Comments

Andre Oosthuizen 21-Jul-20 4:44am

Which variable is not defined? Show us your html code to see what you are posting.

steph99999 21-Jul-20 4:47am

<title>Insert Data

>
>
>
>
User Type User ID Booking Date Starting TimeFacility
Swimming attire rental
Changing area and showers
Personal locker
Sauna and steam room

Payment Type
Please Select
Cash
Debit
Credit Card

>
>
>
>
User Type User ID Booking Date Starting TimeFacility
Swimming attire rental
Changing area and showers
Personal locker
Sauna and steam room

Payment Type
Please Select
Cash
Debit
Credit Card

steph99999 21-Jul-20 4:48am

here's the html code

<title>Insert Data

>
>
>
>
User Type User ID Booking Date Starting TimeFacility
Swimming attire rental
Changing area and showers
Personal locker
Sauna and steam room

Payment Type
Please Select
Cash
Debit
Credit Card

Richard MacCutchan 21-Jul-20 4:54am

You have still not explained which variable is not defined.

steph99999 21-Jul-20 4:56am

all variable

Richard MacCutchan 21-Jul-20 5:17am

Please do not paraphrase. Show us the actual error messages that you see.

Richard Deeming 21-Jul-20 12:35pm

Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation to build a SQL query. ALWAYS use a parameterized query.

PHP: SQL Injection - Manual[^]

2 solutions

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Richard MacCutchan · Answer 1 · 2020-07-20T23:09:00

You are missing the open and close brace for the if statement. But more importantly you are creating local variables in the if(isset, which are not visible to the following code. So it should be:

PHP

if(isset($_POST['submit'])){
    if(!empty($_POST['usertype']) && !empty($_POST['userid']) && !empty($_POST['usertype']) && !empty($_POST['bookingdate']) && !empty($_POST['startingtime']) && !empty($_POST['facility']) && !empty($_POST['payment']) ) { // ***** open brace needed here 
        $usertype=$_POST['usertype'];
        $userid=$_POST['userid'];
        $bookingdate=$_POST['bookingdate'];
        $startingtime=$_POST['startingtime'];
        $facility=$_POST['facility'];
        $payment=$_POST['payment'];
    
        // NB these lines need to be inside the two if blocks
        $sql = "INSERT INTO booking(B_UserType,B_UserID, B_Date,B_StartTime,B_Facility ,B_PaymentType) VALUES ('$usertype','$userid','$bookingdate','$startingtime','$facility','$payment')";
        $run = mysqli_query($conn, $sql ) or die (mysqli_error());
        if($run)
        {
            echo "Inserted";
        }else
        {
            echo "Not inserted";
        }
    } // ***** close brace needed here 
} // end of 'isset' test

steph99999 · Answer 2 · 2020-07-20T22:49:00

HTML

<!DOCTYPE html>
<html>    
    <head>    
        <title>Insert Data</title>    
    </head>   

    <body>    
	<form name= "details" action = "details.php" method ="post">
        <table>

<tr>
<td>User Type </td>
<td><input type = "text" name = "usertype" value =""></td>>
</tr>

<tr>
<td>User ID </td>
<td><input type = "text" name = "userid" value =""></td>>
</tr>

<tr>
<td>Booking Date </td>
<td><input type = "text" name = "bookingdate" value =""></td>>
</tr>

<tr>
<td>Starting Time</td>
<td><input type = "text" name = "startingtime" value =""></td>>
</tr>

<tr>
<td>Facility</td>
<td><select multiple size="4" name ="facility">
          <option value="attire">Swimming attire rental</option>
          <option value="changing">Changing area and showers</option>
          <option value="locker">Personal locker</option>
          <option value="steam">Sauna and steam room</option>
          
</select>
</td>
</tr>

<tr>
<td>Payment Type</td>
<td><select name ="payment">
<option value = "">Please Select</option>
<option value = "cash">Cash</option>
<option value = "debit">Debit</option>
<option value = "credit">Credit Card</option>
</select>
</td>
<tr>
<td><input type = "submit" value="Submit"></td>
</tr>

</table>
</form>
</html>

Notice: undefined variable

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