Error in your SQL syntax

Beginners - Learning PHP
#1

Error description: 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 ‘’ at line 1
Can someone help me find this?

session_start();
//initialize variables
$Naam = "";
$Email = "";
$Pand = "";
$Huisnummer = "";
$Deel = "";
$id = 0;
// connect to database
$db = mysqli_connect("xxx","xxx","xxx","xxx");
//update records
if (isset($_POST['update'])){
    $Naam = mysqli_real_escape_string($_POST['Naam']);
    $Email = mysqli_real_escape_string($_POST['Email']);
    $Pand = mysqli_real_escape_string($_POST['Pand']);
    $Huisnummer = mysqli_real_escape_string($_POST['Huisnummer']);
    $Deel = mysqli_real_escape_string($_POST['Deel']);
    $id = mysqli_real_escape_string($_POST['id']);

    if (!mysqli_query($db, "UPDATE Info SET Naam='$Naam' , Email='$Email' , Pand='$Pand' , Huisnummer='$Huisnummer' , Deel='$Deel' WHERE id=$id")){
        echo("Error description: " . mysqli_error($db));
    }
    header('location: overzichtlocatie.php');
}
// retrieve records
$results = mysqli_query($db, "SELECT * FROM info");
#2

Is the table name, “info” or “Info”? it makes a difference.

#3

Info with the capital letter.

#4

Your second query is using the lower case version. Other than that, I don’t see an issue other than the using variables in the query.

#5

I have changed the info to Info, but still getting the same error…

#6

if you hardcode the values in rather than using variables, does the error still trigger?

#7

What does that mean? How can I do that?

#8 
UPDATE Info SET Naam='Alex' , Email='[email protected]' , Pand='Pand' , Huisnummer='Huisnummer' , Deel='Deel' WHERE id=1
#9

So without the if command?

#10

I just changed the query to show you what I meant. Everything else would stay the same

#11

Still the same error. I will rewrite this again then… Thanks for your help!

#12

Then i would check the column names next.

#13

I have checked the column names, I have used the same names as in the database.

#14

so when you put in the query into something like PHPAdmin does it also fail?

#15 
$db = new mysqli("xxx","xxx","xxx","xxx");
//update records
if (isset($_POST['update'])){
	$sql = "UPDATE Info SET Naam=?, Email=?, Pand=?, Huisnummer=?, Deel=? WHERE id=?";

    $stmt = $conn->prepare($sql);
	$stmt->bind_param("sssssi", $_POST['Naam'], 
								$_POST['Email'], 
								$_POST['Pand'], 
								$_POST['Huisnummer'], 
								$_POST['Deel'], 
								$_POST['id']);
      
    $stmt->execute();
    header('location: overzichtlocatie.php');
}
#16

I have rewritten my code, and it works now! Thankyou for your help!

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