How do I solve these errors? Please help me with this PHP/SQL table issue

Question

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I have a table with a list of people, and I want to have a option where you can edit the information shown of those people listed. I have a editperson.php file and a editsql.php file. I get these errors when trying to submit new values for the persons:

Click here -> https://gyazo.com/5e5e52c26b1b978d7802f58eaafba16d[^]

Here is the code for editperson.php

HTML

<!doctype html>
<html>
<body>
	Skriv inn nye verdier for denne personen
	<form action = 'editsql.php' method = 'POST'>
	<?php
		include('oppkobling.php');

		$id = $_POST['id'];

		// Lag SQL-setning
		$query = 'SELECT * FROM rikestemennesker WHERE id = '.$id;

		// Kjør spørringen
		$resultat = mysqli_query($db, $query);
		$rad = mysqli_fetch_array($resultat);

		//echo "<input type = 'hidden' name = 'editmnr' value = ".$id.">";
		echo "Plass: <input type = 'text' value = '".$rad['id']."' size='30' 
         name='editplass'><br>";
		echo "Navn: <input type = 'text' value = '".$rad['Navn']."' size='30' 
         name='editnavn'><br>";
		echo "Land: <input type = 'text' value = '".$rad['Land']."' size='30' 
         name='editland'><br>";
		echo "Nettoformue: <input type = 'text' value = 
         '".$rad['Nettoformue']."' size='50' name='editformue'><br>";
		echo "Kilde: <input type='text' value = '".$rad['Kilde']."' size ='50' 
         name ='editkilde'><br>";
		// Koble ned
		mysqli_close($db);
		mysqli_free_result($resultat);


	?>
		<input type = 'submit' value = 'Oppdater'>
	</form>
</body>
</html>

and here is the code for editsql.php

PHP

<?php
	include('oppkobling.php');
	$n = $_POST['editnavn'];
	$a = $_POST['editformue'];
  $l = $_POST['editland'];
	$m = $_POST['editplass'];
	$k = $_POST['editkilde'];

	$query = "UPDATE rikestemennesker SET
	navn = '$n',
	land = '$l',
	formue = '$a',
	kilde = '$k',
	WHERE plass = $m";

	mysqli_query($db, $query);

	mysqli_close($db);
	header("Location: index.php");
?>

What I have tried:

I have tried to look over all variables, and checked that the code is linked to the database

Posted 21-Jan-18 9:28am

Member 13629755

Updated 21-Jan-18 15:34pm

Patrice T

v3

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Patrice T · Answer 1 · 2018-01-21T12:39:00

PHP

$query = 'SELECT * FROM rikestemennesker WHERE id = '.$id;
$query = "UPDATE rikestemennesker SET
navn = '$n', land = '$l', formue = '$a', kilde = '$k',
WHERE plass = $m";

Not a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]

Mohibur Rashid · Answer 2 · 2018-01-21T15:34:00

There is a comma before WHERE clause in your update query.
since you are using mysqli, then use mysqli class[^] instead

and rewrite your select query as

PHP

$query = 'SELECT * FROM rikestemennesker WHERE id = ?';
$stmt = $mysqli->prepare("id", $id);
$mysqli->execute($db, $query);
$row = $result->get_result()->fetch_assoc();// add some verification

Do similar with update command too.

How do I solve these errors? Please help me with this PHP/SQL table issue

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