我该如何解决这些错误？请帮我解决这个PHP / SQL表问题

问题描述：

我有一张包含人员列表的表格，我想有一个选项，您可以在其中编辑所列人员的信息。我有一个editperson.php文件和一个editsql.php文件。我在尝试为这些人提交新值时遇到这些错误：

点击此处 - > https://gyazo.com/5e5e52c26b1b978d7802f58eaafba16d [ ^ ]

这是editperson.php的代码

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

<!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>

和这里是editsql.php的代码

and here is the code for editsql.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");
?>

我有什么试过：

我试图查看所有变量，并检查代码是否链接到数据库

What I have tried:

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

答

id =

_POST [' ID 跨度>];

// 滞后SQL设置

_POST['id']; // Lag SQL-setning

query = ' SELECT * FROM rikestemennesker WHERE id ='。

我该如何解决这些错误？请帮我解决这个PHP / SQL表问题

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