Quote:
I need help with SQL syntax error.
1 of your problems is that you can't know what is the query.
I would start by replace
if (!mysqli_query($db,"UPDATE Info SET Naam= '$Naam' , Email= '$Email' , Pand= '$Pand' , Huisnummer= '$Huisnummer' , Deel= '$Deel' WHERE id=$id")){
by something like
$Query= "UPDATE Info SET Naam= '$Naam' , Email= '$Email' , Pand= '$Pand' , Huisnummer= '$Huisnummer' , Deel= '$Deel' WHERE id=$id";
if (!mysqli_query($db,$Query)){
This little change allow you to print $Query or inspect it with debugger. This way, you can know what was your real query.
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Your code do not behave the way you expect, or you don't understand why !
There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.
The downside of this solution:
Debugger - Wikipedia, the free encyclopedia[
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Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[
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Basic Debugging with Visual Studio 2010 - YouTube[
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phpdbg | php debugger[
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Debugging techniques for PHP programmers[
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The debugger is here to only show you what your code is doing and your task is to compare with what it should do.